Class 12 NCERT Solutions for Physics Chapter 1

Q.1 What is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air?

Ans.

\begin{array}{l}\text{Here,}c\text{harge on first sphere},{\text{q}}_{\text{1}}=\text{2}\times \text{1}{0}^{-\text{7}}\text{C}\\ \text{Charge on second sphere},{\text{q}}_{\text{2}}=\text{3}\times \text{1}{0}^{-\text{7}}\text{C}\\ \text{Distance between spheres},\text{r}=\text{3}0\text{cm}\\ =0.\text{3 m}\\ \text{Electrostatic force between the spheres is given as:}\\ \text{F=}\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}^2}\\ \text{Here},{\in }_0=\text{Permittivity of free space}\\ \frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^{-2}\\ \therefore F=\frac{9\times {10}^9\times \text{2}\times \text{1}{0}^{-\text{7}}\times \text{3}\times \text{1}{0}^{-\text{7}}}{{\left(0.3\right)}^2}\\ =6\times {10}^{-3}N\\ \therefore \text{Force between the two small charged spheres}\\ \text{=6}\times \text{1}{0}^{-\text{3}}\text{N}.\\ \text{Since}t\text{he charges are}\text{of same nature,}\\ \therefore \text{Force between them will be repulsive}.\end{array}

Q.2 The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{Electrostatic force on first sphere},\text{F}=0.\text{2 N}\\ \text{Charge on first sphere},{\text{q}}_{\text{1}}=0.\text{4}\mu \text{C}\\ =0.\text{4}\times \text{1}{0}^{-\text{6}}\text{C}\\ \text{Charge on second sphere},{\text{q}}_{\text{2}}=-0.\text{8}\mu \text{C}\\ =-0.\text{8}\times \text{1}{0}^{-\text{6}}\text{C}\\ \text{Electrostatic force between the spheres is given as:}\\ \text{F=}\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}^2}\\ \text{Here},{\epsilon }_0=\text{Permittivity of free space}\\ \frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^{-2}\\ \therefore r^2=\frac{q_1{q}_2}{4\pi {\epsilon }_{0}F}\\ =\frac{9\times {10}^9\times 0.4\times \text{1}{0}^{-6}\times 0.8\times \text{1}{0}^{-6}}{0.2}\\ =144\times {10}^{-4}N\\ r=\sqrt{144\times {10}^{-4}}=0.12m\\ \therefore \text{Distance between the two spheres =}0.12\text{m}.\\ \left(\text{b}\right)\text{As}b\text{oth the spheres attract each other with}\\ \text{same force,}\\ \therefore F\text{orce on}\text{the second sphere due to the first sphere}\\ \text{is}0.\text{2 N}.\end{array}

Q.3 Check that the ratio ke²/Gm_em_p is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Ans.

\begin{array}{l}\text{The given ratio =}\frac{k{e}^2}{G{m}_e{m}_p}\\ \text{Here},\text{G}\text{is}\text{the Gravitational constant,}its{\text{unit=Nm}}^{\text{2}}{\text{kg}}^{-\text{2}}\\ {\text{m}}_{\text{e}}{\text{and m}}_{\text{p}}\text{are}\text{masses of electron and proton,}\text{their}\\ \text{unit= kg}.\\ \text{e is}\text{the electric charge,}i\text{ts unit= C}\\ \text{K=}\text{A}\text{constant=}\frac{1}{4\pi {\epsilon }_0}\\ {\epsilon }_0\text{is}thep\text{ermittivity of free space,}i\text{ts unit=}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\\ \therefore Thegivenratio,\frac{k{e}^2}{G{m}_e{m}_p}=\frac{\left[{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\right]\left[{\text{C}}^{-\text{2}}\right]}{\left[{\text{Nm}}^{\text{2}}{\text{kg}}^{-\text{2}}\right]\left[\text{kg}\right]\left[\text{kg}\right]}\\ =M^0{L}^0{T}^0\\ \therefore \text{The given ratio is dimensionless}.\\ \text{e = 1}{\text{.6 × 10}}^{\text{-19}}\text{C}\\ \text{G = 6}{\text{.67 × 10}}^{\text{-11}}{\text{N m}}^{\text{2}}{\text{kg}}^{\text{-2}}\\ {\text{m}}_{\text{e}}\text{= 9}{\text{.1 × 10}}^{\text{-31}}\text{kg}\\ {\text{m}}_{\text{p}}\text{= 1}{\text{.66 × 10}}^{\text{-27}}\text{kg}\\ \therefore N\text{umerical value of the given ratio}\text{is}\text{given}\text{as:}\\ \frac{k{e}^2}{G{m}_e{m}_p}=\frac{9\times {10}^9\times {\left(\text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\right)}^2}{\text{6}.\text{67}\times \text{1}{0}^{-\text{11}}\times \text{9}.\text{1}\times \text{1}{0}^{-\text{31}}\times \text{1}.\text{66}\times \text{1}{0}^{-\text{27}}}\\ \approx 2.3\times {10}^{39}\\ \text{This is the ratio of electrostatic force to gravitational}\\ \text{force between an}\text{electron and a proton,}\text{keeping}\\ \text{distance between them constant}.\end{array}

Q.4 (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Ans.

(a) Electric charge of a body is quantized means that only integral (1, 2, …., n) number of elementary charge can be transferred from one body to the other. The elementary charge is charge on an electron or a proton which cannot be transferred in fraction.
(b) At macroscopic level, the charges used are enormous as compared to the magnitude of electric charge. Thus, quantization of charge can be ignored and can be considered as continuous charge.

Q.5 When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Ans.

When a glass rod is rubbed with a silk cloth, charges of equal magnitude but of opposite nature appear on the two bodiesas the charges are created in pairs. The algebraic sum of charges produced on the two rubbed bodies is zero. The net charge on the two bodies before rubbing was also zero. Therefore, this phenomenon is consistent with the law of conservation of energy. The similar phenomenon is observed with many other pairs of bodies.

Q.6 Four point charges q_A = 2 μC, q_B = −5 μC, q_C = 2 μC, and q_D = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

Ans.

\begin{array}{l}\text{In}\text{the given figure,}\text{there}\text{is}\text{a}\text{square of side 1}0\text{cm with}\\ \text{four charges placed at its corners}.\\ \text{S is the centre of the square}.\\ \text{where},\\ \text{Length}\text{of}\text{side}\text{of}\text{square=AB}=\text{BC}=\text{CD}=\text{AD}=\text{1}0\text{cm}\\ \text{Length}\text{of}\text{diagonal}\text{of}\text{square=AC}=\text{BD}=\text{10}\sqrt{2}\text{cm}\\ \text{AS}=\text{SC}=\text{DS}=\text{SB}=5\sqrt{2}\text{cm}\\ \text{A charge of amount 1}\mu \text{C is placed at the}centreSof\\ thesquare.\end{array}

Force of repulsion on 1 μC charge because of 2 μC charges at A and C are equal in magnitude but opposite in direction. Therefore, they cancel each other.
Similarly, force of attraction on 1 μC charge because of -5 μC charges at A and C are equal in magnitude but opposite in direction. Therefore, they also cancel each other.
Therefore, net force on 1 μC charge at centre S is zero.

Q.7 (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point?

Ans.

(a) An electrostatic field line is a continuous curve because it represents the actual path of the unit positive charge which experiences a continuous force in an electrostatic field. It cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.
(b) If two field lines cross each other at a point, then electric field will show two directions at that point. This is impossible. Therefore, two field lines never cross each other at any point.

Q.8 Two point charges q_A = 3 μC and q_B = −3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10⁻⁹ C is placed at this point, what is the force experienced by the test charge?

Ans.

\begin{array}{l}\left(\text{a}\right)\text{The situation}\text{of}\text{the}\text{problem is shown in the given}\\ \text{figure}.\\ \text{Here,}\text{O is the mid}-\text{point of line AB}.\\ {\text{q}}_{\text{A}}\text{=3μC}\\ {\text{=3×10}}^{\text{-6}}\text{C}\\ {\text{q}}_{\text{B}}\text{=-3μC}\\ {\text{=-3×10}}^{\text{-6}}\text{C}\\ \text{Distance between two charges},\text{AB}=\text{2}0\text{cm}\\ \therefore \text{AO}=\text{OB}\\ =\text{1}0\text{cm}\\ \text{Net electric field at O}=\text{E}\\ \text{Electric field at point O caused by}+\text{3}\mu \text{C charge}\text{is}\\ \text{given}\text{as:}\\ {\text{E}}_{\text{1}}=\frac{q_A}{4\pi {\epsilon }_0{\left(AO\right)}^2}\\ \text{=}\frac{3\times {10}^{-6}}{4\pi {\epsilon }_0{\left(\text{1}0\times {10}^{-2}\right)}^2}N{C}^{-1}alongOB\\ \text{Where},{\epsilon }_0=\text{Permittivity of free space}\\ \frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^{-2}\\ \text{Magnitude of electric field at point O caused by}-\text{3}\mu \text{C}\\ \text{charge}\text{is}\text{given}\text{as:}\\ {\text{E}}_2=\left|\frac{q_B}{4\pi {\epsilon }_0{\left(OB\right)}^2}\right|\\ =\left|\frac{-3\times {10}^{-6}}{4\pi {\epsilon }_0{\left(\text{1}0\times {10}^{-2}\right)}^2}\right|\\ \text{=}\frac{3\times {10}^{-6}}{4\pi {\epsilon }_0{\left(\text{1}0\times {10}^{-2}\right)}^2}N{C}^{-1}alongOB\\ \therefore E={\text{E}}_{\text{1}}+{\text{E}}_{\text{2}}\\ =\frac{3\times {10}^{-6}}{4\pi {\epsilon }_0{\left(\text{1}0\times {10}^{-2}\right)}^2}+\frac{3\times {10}^{-6}}{4\pi {\epsilon }_0{\left(\text{1}0\times {10}^{-2}\right)}^2}\\ =2\times \left[\left(9\times {10}^9\right)\times \frac{3\times {10}^{-6}}{{\left(\text{1}0\times {10}^{-2}\right)}^2}\right]\\ =5.4\times {10}^{6}N{C}^{-1}\text{along OB}\\ {\text{E}}_{\text{2}}=\text{5}.\text{4}\times \text{1}{0}^{\text{6}}{\text{NC}}^{\text{-1}}\text{along OB}\\ \therefore \text{The}\text{electric field at mid}-\text{point O is 5}.\text{4}\times \text{1}{0}^{\text{6}}{\text{NC}}^{-\text{1}}\\ \text{along OB}.\\ \left(\text{b}\right)\text{A test charge of magnitude 1}.\text{5}\times \text{1}{0}^{-\text{9}}\text{C is placed}\\ \text{at}\text{mid}-\text{point O}.\\ \text{Here,}\text{q}=\text{1}.\text{5}\times \text{1}{0}^{-\text{9}}\text{C}\\ \text{Let}f\text{orce experienced by the test charge}=\text{F}\\ \therefore \text{F}=\text{qE}\\ \text{= 1}{\text{.5×10}}^{\text{-9}}\text{×5}{\text{.4×10}}^{\text{6}}\text{= 8}{\text{.1×10}}^{\text{-3}}\text{N;}\text{along BA}\text{.}\end{array}

Q.9 A system has two charges q_A = 2.5 × 10⁻⁷ C and q_B = −2.5 × 10⁻⁷ C located at points A: (0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?

Ans.

\begin{array}{l}{\text{The charges q}}_{\text{A}}\text{and}{\text{q}}_{\text{B}}\text{can be located in a coordinate}\\ \text{frame of reference as shown in the given figure}\text{.}\\ \text{Magnitude of charge}\text{at}{\text{A, q}}_{\text{A}}\text{= 2}{\text{.5 × 10}}^{\text{-7}}\text{C}\\ \text{Magnitude of charge}\text{at}{\text{B, q}}_{\text{B}}\text{= -2}{\text{.5 × 10}}^{\text{-7}}\text{C}\\ \text{Total charge of the system,}{\text{q = q}}_{\text{A}}{\text{+ q}}_{\text{B}}\\ \text{= 2}{\text{.5 × 10}}^{\text{7}}\text{C – 2}{\text{.5 × 10}}^{\text{-7}}\text{C}\\ \text{= 0}\text{C}\\ \text{Distance between the}\text{two charges,}\text{d=15+15}\\ \text{=30 cm}\\ \text{=0}\text{.3 m}\\ \text{Electric dipole moment of the system is given as:}\\ \text{p}={\text{q}}_{\text{A}}\times \text{d}\\ ={\text{q}}_{\text{B}}\times \text{d}\\ =\text{2}.\text{5}\times \text{1}{0}^{-\text{7}}\times 0.\text{3}\\ =\text{7}.\text{5}\times \text{1}{0}^{-\text{8}}\text{C m along positive z}-\text{axis}\\ \therefore \text{Electric dipole moment of the system is}\text{7}.\text{5}\times \text{1}{0}^{-\text{8}}\text{Cm}\\ \text{along positive}\text{z}-\text{axis}.\end{array}

Q.10 An electric dipole with dipole moment 4 × 10⁻⁹ C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ N C⁻¹. Calculate the magnitude of the torque acting on the dipole.

Ans.

\begin{array}{l}\text{Here,}{\text{electric dipole moment, p = 4×10}}^{\text{-9}}\text{Cm}\\ \text{Angle made by p with the}\text{uniform electric field, θ=30°}\\ \text{Magnitude}\text{of}{\text{electric field, E = 5×10}}^{\text{4}}{\text{NC}}^{\text{-1}}\\ \text{Torque acting on the dipole is given by,}\\ \text{τ = pE sinθ}\\ {\text{=4×10}}^{\text{-9}}\text{×}{\text{5×10}}^{\text{4}}\text{sin30°}\\ {\text{=20×10}}^{\text{-5}}\text{×}\frac{\text{1}}{\text{2}}\\ {\text{=10}}^{\text{-4}}\text{Nm}\\ \text{The}\text{magnitude of the torque acting on the dipole is}\\ {\text{10}}^{\text{-4}}\text{Nm}\text{.}\end{array}

Q.11 A polythene piece rubbed with wool is found to have a negative charge of 3 × 10⁻⁷ C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here, charge on the polythene piece,}\\ {\text{q=-3×10}}^{\text{-7}}\text{C}\\ \text{Charge on an electron, e = -1}{\text{.6×10}}^{\text{-19}}\text{C}\\ \text{Let}\text{number of electrons transferred to polythene}\\ \text{piece}\text{from}\text{wool= n}\\ \text{We}\text{can}\text{calculate}\text{n}\text{by using the relation,}\\ \text{q = ne}\\ \text{n=}\frac{\text{q}}{\text{e}}\text{=}\frac{{\text{-3×10}}^{\text{-7}}}{\text{-1}{\text{.6×10}}^{\text{-19}}}\\ \text{= 1}{\text{.87×10}}^{\text{12}}\\ \text{Number of electrons transferred from wool to}\\ \text{polythene is 1}{\text{.87 ×10}}^{\text{12}}\text{.}\\ \left(\text{b}\right)\text{Yes,}\text{transfer of mass takes place}\text{from}\text{wool}\text{to}\\ \text{the}\text{polythene}\text{piece}\text{.}\\ \text{As}\text{mass}\text{of}\text{electron,}{\text{m}}_{\text{e}}\text{= 9}{\text{.1×10}}^{\text{-31}}\text{kg}\\ \text{Total mass transferred from}\text{wool}\text{to}\text{the}\text{polythene,}\text{m}\\ {\text{= m}}_{\text{e}}\text{× n}\\ \text{= 9}{\text{.1×10}}^{\text{-31}}\text{×1}{\text{.85×10}}^{\text{12}}\\ \text{= 1}{\text{.706×10}}^{\text{-18}}\text{kg}\\ \therefore A\text{negligible amount of mass is transferred from}\\ \text{wool to polythene}.\end{array}

Q.12 (a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10⁻⁷ C? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,}{\text{charge on sphere A, q}}_{\text{A}}\text{=6}{\text{.5×10}}^{\text{-7}}\text{C}\\ {\text{Charge on sphere B, q}}_{\text{B}}\text{=6}{\text{.5×10}}^{\text{-7}}\text{C}\\ \text{Distance between spheres, r=50 cm}\\ \text{=0}\text{.5 m}\\ \text{Force of electrostatic}\text{repulsion between the two}\\ \text{spheres}\text{is}\text{given}\text{as:}\\ \text{F=}\frac{{\text{q}}_{\text{A}}{\text{q}}_{\text{B}}}{4\pi {\epsilon }_0{r}^2}\\ \text{Here},{\epsilon }_0=P\text{ermittivity}\text{of}f\text{ree space}=\text{9}\times \text{1}{0}^{\text{9}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\\ \therefore \text{F=}\frac{\text{9}\times \text{1}{0}^{\text{9}}\times {\left(\text{6}.\text{5}\times \text{1}{0}^{-\text{7}}\right)}^2}{{\left(0.\text{5}\right)}^2}\\ \text{= 1}{\text{.52 × 10}}^{\text{-2}}\text{N}\\ \therefore \text{The}\text{force between the two spheres is 1}.\text{52}\times \text{1}{0}^{-\text{2}}\text{N}.\\ \left(\text{b}\right)\text{After doubling the amount}\text{of}\text{charge},\\ {\text{Charge on sphere A, q}}_{\text{A}}\text{= Charge on sphere B,}{\text{q}}_{\text{B}}\\ =\text{2}\times \text{6}.\text{5}\times \text{1}{0}^{-\text{7}}\text{C}\\ =\text{1}.\text{3}\times \text{1}{0}^{-\text{6}}\text{C}\\ \text{As}\text{distance between the spheres is halved,}\\ \therefore r=\frac{0.\text{5}}{2}\\ =0.25m\\ \text{Force of repulsion between the two spheres}\text{is}\text{given}\text{as},\\ \text{F=}\frac{{\text{q}}_{\text{A}}{\text{q}}_{\text{B}}}{4\pi {\epsilon }_0{r}^2}\\ \text{=}\frac{{\text{9×10}}^{\text{9}}\text{×1}{\text{.3×10}}^{\text{-6}}\text{×1}{\text{.3×10}}^{\text{-6}}}{{\left(\text{0}\text{.25}\right)}^{\text{2}}}\\ \text{= 16×1}{\text{.52×10}}^{\text{-2}}\\ \text{= 0}\text{.243 N}\\ \therefore F\text{orce between the two spheres is}0.\text{243 N}.\end{array}

Q.13 Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Ans.

\begin{array}{l}\text{Here,}\text{initial charge on each sphere,}\\ \text{q = 6}{\text{.5 × 10}}^{\text{-7}}\text{C}\\ \text{Distance between the spheres A and B, r = 0}\text{.5 m}\\ \text{When sphere A is touched with an uncharged sphere C},\\ theirch\arg esareequallyshared.\\ \therefore \text{Charge}\text{left}\text{on}\text{sphere}\text{A,}{\text{q}}_{\text{1}}\text{=}\frac{\text{6}{\text{.5 × 10}}^{\text{-7}}}{\text{2}}\\ =3.25\times \text{1}{0}^{-\text{7}}C\\ \text{When sphere C with charge is brought in contact with}\\ \text{sphere B with charge q},theirch\arg esareshared\\ equally.\\ \therefore \text{Charge}\text{left}\text{on}\text{sphere}\text{B,}{\text{q}}_{\text{2}}\text{=}\frac{\left(\text{6}\text{.5+3}\text{.25}\right){\text{×10}}^{\text{-7}}}{\text{2}}\text{C}\\ =4.875\times \text{1}{0}^{-\text{7}}C\\ Asforceofrepulsionisgivenbytherelation,\\ F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}\\ Here,\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^{-2}\\ \therefore \text{F=}\frac{{\text{9×10}}^{\text{9}}\text{×3}{\text{.25×10}}^{\text{-7}}\text{×4}{\text{.875×10}}^{\text{-7}}}{{\left(\text{0}\text{.5}\right)}^{\text{2}}}\\ \text{=5}{\text{.7×10}}^{\text{-3}}\text{N}\\ \therefore \text{The force of attraction between the two spheres is}\\ \text{5}.\text{7}0\text{3}\times \text{1}{0}^{-\text{3}}\text{N}.\end{array}

Q.14 Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field.
Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Ans.

As opposite charges attract each other and same charges repel each other. Therefore, particle 1 and particle 2 are negatively charged and particle 3 is positively charged.
The displacement or the amount of deflection for a given velocity is directly proportional to the charge to mass ratio (emf). As the deflection of particle 3 is the maximum, therefore it has the highest charge to mass ratio.

Q.15

\begin{array}{l}{\text{Consider a uniform electric field E = 3×10}}^{\text{3}}\widehat{i}{\text{NC}}^{\text{-1}}\text{.}\\ \text{(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?}\\ \text{(b) What is the flux through the same}\text{square if the normal to its plane makes a 60}°\text{angle with the x-axis?}\end{array}

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{electric field,}\overrightarrow{\text{E}}{\text{= 3×10}}^{\text{3}}\widehat{i}{\text{NC}}^{\text{-1}}\\ \text{Magnitude of electric field,}\left|\overrightarrow{\text{E}}\right|{\text{= 3×10}}^{\text{3}}{\text{NC}}^{\text{-1}}\\ \text{Side of square, s = 10 cm = 0}\text{.1 m}\\ {\text{Area of square, A = s}}^{\text{2}}\text{= 0}{\text{.01 m}}^{\text{2}}\\ \text{Since}\text{the}\text{plane of the square is parallel to the y-z plane}\\ \therefore \text{Angle between the unit}\text{vector normal to the plane and electric field, θ = 0}°\text{.}\end{array}

Q.16 What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Ans.

Since all the faces of a cube are parallel to the coordinate axes, therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. Therefore, net flux through the cube is zero.

Q.17 Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10³ N m²C^-1.
(a) What is the net charge inside the box?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,}n\text{et outward flux through the surface of the}\\ \text{box},\varphi =\text{8}.0\times \text{1}{0}^{\text{3}}{\text{N m}}^{\text{2}}{\text{C}}^{\text{-1}}\\ \text{For a body containing net charge q},\text{flux is given}\text{as:}\\ \varphi =\frac{q}{{\epsilon }_0}\\ Here,{\epsilon }_0=\text{Permittivity of free space}\\ =\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}{\text{N}}^{-\text{1}}{\text{C}}^{\text{2}}{\text{m}}^{-\text{2}}\\ \text{q}={\epsilon }_0\varphi \\ =\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}\times \text{8}.0\times \text{1}{0}^{\text{3}}\\ =\text{7}.0\text{8}\times \text{1}{0}^{-\text{8}}\\ =0.0\text{7}\mu \text{C}\\ \therefore \text{Net charge inside the box is}0.0\text{7}\mu \text{C}.\\ \left(\text{b}\right)\text{If net flux through}\text{the}\text{surface}\text{is zero},\text{then as per}\\ \text{gauss theorem it can be concluded that net charge inside}\\ \text{the body is zero}.\text{In}\text{such}\text{situation,}t\text{he body may have}\\ \text{equal amount of positive and negative charges}.\end{array}

Q.18 A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

Ans.

The square of side 10 cm can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem, total electric flux through all the six faces of the cube is given as:

\begin{array}{l}{\varphi }_{total}=\frac{q}{{\epsilon }_0}\\ \therefore \text{Electric flux through the square},\varphi =\frac{{\varphi }_{total}}{6}\\ =\frac{1}{6}\frac{q}{{\epsilon }_0}\\ \text{Here},{\epsilon }_0=\text{Permittivity of free space}\\ =\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}{\text{N}}^{-\text{1}}{\text{C}}^{\text{2}}{\text{m}}^{-\text{2}}\\ \text{Point}\text{charge,}\text{q}=\text{1}0\mu \text{C}\\ =\text{1}0\times \text{1}{0}^{-\text{6}}\text{C}\\ \therefore {\varphi }_E=\frac{1}{6}\times \frac{\text{1}0\times \text{1}{0}^{-\text{6}}}{\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}}\\ \text{= 1}.\text{88}\times \text{1}{0}^{\text{5}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{1}}\\ \therefore \text{Electric flux through the square is 1}.\text{88}\times \text{1}{0}^{\text{5}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{1}}.\end{array}

Q.19 A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Ans.

\begin{array}{l}Here,\text{Net charge inside the cube},q=\text{2}.0\mu \text{C}\\ =\text{2}\times \text{1}{0}^{-\text{6}}\text{C}\\ \text{Net electric flux}\left({\varphi }_{\text{net}}\right)\text{through the cubic surface is}\\ \text{given by}\text{the}\text{relation},\\ {\varphi }_{\text{net}}=\frac{q}{{\epsilon }_0}\\ \text{Here},{\epsilon }_0=\text{Permittivity of free space}\\ =\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}{\text{N}}^{-\text{1}}{\text{C}}^{\text{2}}{\text{m}}^{-\text{2}}\\ \therefore {\varphi }_{\text{net}}=\frac{\text{2}\times \text{1}{0}^{-\text{6}}}{\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}}\\ =\text{2}.\text{26}\times \text{1}{0}^{\text{5}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{1}}\\ \therefore \text{Net electric flux through the surface is}\\ \text{2}.\text{26}\times \text{1}{0}^{\text{5}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{1}}.\end{array}

Q.20 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ NC^-1 and points radially inward, what is the net charge on the sphere?

Ans.

\begin{array}{l}Here,radiusofthesphere,r=10cm\\ \text{Electric field}\text{intensity},E=-\text{1}.\text{5}\times \text{1}{0}^{\text{3}}{\text{NC}}^{\text{-1}}\\ \left(Negativesignforinwardfield\right)\\ \text{Distance from the centre}ofthesphere,d=\text{2}0\text{cm}\\ =0.\text{2 m}\\ \text{Let}\text{net}\text{charge}\text{on}\text{the}\text{sphere=q}\\ \text{Electric field intensity}\text{is given by the relation},\\ \because \text{d > r}\\ E=\frac{q}{4\pi {\epsilon }_0{d}^2}\\ Here,{\epsilon }_0=\text{Permittivity of free space}\\ =\text{9}\times \text{1}{0}^{\text{9}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\\ \therefore q=E\times 4\pi {\epsilon }_0{d}^2\\ =\frac{\text{-1}.\text{5}\times \text{1}{0}^{\text{3}}\times {\left(0.\text{2}\right)}^2}{\text{9}\times \text{1}{0}^{\text{9}}}\\ =-\text{6}.\text{67}\times \text{1}{0}^{\text{9}}\text{C}\\ =-\text{6}.\text{67 nC}\\ \therefore \text{Net charge on the sphere is -6}.\text{67 nC}\text{.}\end{array}

Q.21 A point charge causes an electric flux of −1.0 × 10³ Nm²C^-1 to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Here,}e\text{lectric flux},\varphi =-\text{1}.0\times \text{1}{0}^{\text{3}}{\text{N m}}^{\text{2}}{\text{C}}^{\text{-1}}\\ \text{Radius of the}\text{Gaussian surface},R=\text{1}0.0\text{cm}\\ \text{Electric flux passing through a surface depends on the}\\ \text{net charge enclosed inside a body}\text{and}i\text{t does not}\\ \text{depend on the size of the body}.\\ \therefore \text{If the radius}\text{of the}\text{Gaussian surface is doubled},\text{then}\\ \text{the flux passing through the surface remains the}\text{same}\text{.}\\ \left(\text{b}\right)\text{Electric flux is given as:}\\ \varphi \text{=}\frac{q}{{\epsilon }_0}\\ \text{Here},\text{q}=\text{Net charge enclosed by the surface}\\ {\epsilon }_0=\text{Permittivity of free space}\\ =\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}{\text{N}}^{-\text{1}}{\text{C}}^{\text{2}}{\text{m}}^{-\text{2}}\\ \therefore q=\varphi {\epsilon }_0=-\text{1}.0\times \text{1}{0}^{\text{3}}\times \text{8}.\text{854}\times \text{1}{0}^{-\text{12}}\\ =-\text{8}.\text{854}\times \text{1}{0}^{-\text{9}}\text{C}=-\text{8}.\text{854 nC}\\ \therefore \text{The value of the point charge is}-\text{8}.\text{854 nC}.\end{array}$

Q.22 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μCm^-2.
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere?

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,}d\text{iameter of sphere},\text{d}=\text{2}.\text{4 m}\\ \therefore \text{Radius of sphere},\text{r}=\frac{\text{d}}{2}\\ \text{= 1}.\text{2 m}\\ \text{Surface charge density}\text{of}\text{sphere},\sigma =\text{8}0.0\mu {\text{Cm}}^{\text{-2}}\\ =\text{8}0\times \text{1}{0}^{-\text{6}}{\text{Cm}}^{\text{-2}}\\ \text{Total charge on the surface of the sphere}\text{is}\text{given}\text{as:}\\ \text{Q}=\sigma \times 4\pi r^2\\ =\text{8}0\times \text{1}{0}^{-\text{6}}\times \text{4}\times \text{3}.\text{14}\times {\left(\text{1}.\text{2}\right)}^{\text{2}}\\ =\text{1}.\text{447}\times \text{1}{0}^{-\text{3}}\text{C}\\ \therefore \text{Charge on the sphere is 1}.\text{447}\times \text{1}{0}^{-\text{3}}\text{C}.\\ \left(\text{b}\right)Here,Q=\text{1}.\text{447}\times \text{1}{0}^{-\text{3}}\text{C}\\ \text{Total electric flux leaving the surface of a sphere is}\\ \text{given by the relation},\\ {\varphi }_{total}=\frac{\text{Q}}{{\epsilon }_0}\\ \text{Here},{\epsilon }_0=\text{Permittivity of free space}\\ =\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}{\text{N}}^{-\text{1}}{\text{C}}^{\text{2}}{\text{m}}^{-\text{2}}\\ \therefore {\varphi }_{total}=\frac{\text{1}.\text{447}\times \text{1}{0}^{-\text{3}}}{\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}}\\ =\text{1}.\text{63}\times \text{1}{0}^{\text{8}}{\text{NC}}^{-\text{1}}{\text{m}}^{\text{2}}\\ \therefore \text{The total electric flux leaving the surface of the}\\ \text{sphere is 1}.\text{63}\times \text{1}{0}^{\text{8}}{\text{NC}}^{-\text{1}}{\text{m}}^{\text{2}}.\end{array}

Q.23 An infinite line charge produces a field of 9 × 10⁴ NC^-1 at a distance of 2 cm. Calculate the linear charge density.

Ans.

$\begin{array}{l}\\ Here,\\ \text{Electric field},\\ E=\text{9}\times \text{1}{0}^{\text{4}}{\text{NC}}^{\text{-1}}\\ Dis\tan ce,\\ d=2cm\\ =0.02m\\ \text{Electric field produced by the infinite line charges at a}\\ \text{distance d having linear charge density}\lambda \text{is given as:}\\ \text{E=}\frac{\lambda }{2\pi {\epsilon }_{0}d}\\ \therefore \lambda =2\pi {\epsilon }_{0}d\times \text{E}\\ Here,\\ {\epsilon }_0=\text{Permittivity of free space}\\ \frac{1}{4\pi {\epsilon }_0}=\text{9}\times \text{1}{0}^{\text{9}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\\ \therefore \lambda =\frac{0.02\times \text{9}\times \text{1}{0}^{\text{4}}}{2\times \text{9}\times \text{1}{0}^{\text{9}}}\\ =\text{1}{0}^{-7}{\text{Cm}}^{\text{-1}}\\ \therefore \text{The linear charge density is}0.1\mu {\text{Cm}}^{\text{-1}}.\end{array}$

Q.24

$\begin{array}{l}\text{Two large, thin metal plates are parallel and close to each other}\text{. On their inner faces,}\\ \text{the plates have surface charge densities of opposite signs and of magnitude}\\ \text{17}{\text{.0×10}}^{\text{-22}}{\text{Cm}}^{\text{-2}}\text{. What is}\overrightarrow{E}\text{:}\left(\text{a}\right)\text{in the}\text{outer region of the first plate,}\left(\text{b}\right)\text{in the outer}\\ \text{region of the second plate, and}\left(\text{c}\right)\text{between the plates?}\end{array}$

Ans.

The situation of the problem is represented in the given figure.

Here, A and B are two parallel plates close to each other. Outer region of plate A is labeled as I, the region between the plates, A and B, is labelled as II and outer region of plate B is labelled as III.

\begin{array}{l}\text{Here,}\text{surface}\text{charge density, σ = 17}{\text{.0×10}}^{\text{-22}}{\text{Cm}}^{\text{-2}}\\ \text{Since}\text{charge is not}\text{enclosed by the respective plates}\text{in the regions, I and III,}\\ \therefore \text{Electric field, E = 0,}\text{in}\text{regions}\text{I and III}\text{.}\\ \text{Thus,}\left(\text{a}\right)\text{and}\left(\text{b}\right)\text{have net electric field zero}\text{.}\\ \\ \left(\text{c}\right)\text{Electric field E in region II is given by,}\\ \text{E =}\frac{\text{σ}}{{\text{ε}}_{\text{0}}}\\ \text{Here,}{\text{ε}}_{\text{0}}\text{=8}{\text{.854×10}}^{\text{-12}}{\text{N}}^{\text{-1}}{\text{C}}^{\text{2}}{\text{m}}^{\text{-2}}\\ \therefore \text{E =}\frac{\text{17}{\text{.0×10}}^{\text{-22}}{\text{Cm}}^{\text{-2}}}{\text{8}{\text{.854×10}}^{\text{-12}}{\text{N}}^{\text{-1}}{\text{C}}^{\text{2}}{\text{m}}^{\text{-2}}}\text{= 1}{\text{.92×10}}^{\text{-10}}{\text{NC}}^{\text{-1}}\\ \therefore \text{Electric field between the plates is 1}{\text{.92×10}}^{\text{-10}}{\text{NC}}^{\text{-1}}\text{.}\end{array}

Q.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ NC⁻¹ in Millikan’s oil drop experiment. The density of the oil is 1.26 gcm⁻³. Estimate the radius of the drop. (g = 9.81 ms⁻²; e = 1.60 × 10⁻¹⁹ C).

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{e}\text{xcess electrons on an oil drop},\text{n = 12}\\ \text{Magnitude\hspace{0.17em}of\hspace{0.17em}Electric field},{\text{E = 2.55×10}}^{\text{4}}{\text{NC}}^{\text{-1}}\\ \text{Density of oil},{\text{ρ = 1.26 gmcm}}^{\text{-3}}\\ {\text{= 1.26 × 10}}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ {\text{Acceleration due to gravity, g = 9.81 m s}}^{\text{-2}}\\ {\text{Charge on an electron, e = 1.6×10}}^{\text{-19}}\text{C}\\ \text{Let\hspace{0.17em}radius of\hspace{0.17em}oil drop = r}\\ \text{As\hspace{0.17em}the\hspace{0.17em}oil\hspace{0.17em}drop\hspace{0.17em}is\hspace{0.17em}stationary,}\\ \therefore \text{Force due to electric field = Weight of the oil drop}\\ \text{F = W}\\ \text{As\hspace{0.17em}F=Eq\hspace{0.17em}and\hspace{0.17em}W=mg}\\ \therefore \text{Eq = mg}\to \text{(i)}\mathrm{}\\ \text{Charge on oil drop,\hspace{0.17em}q= ne→(ii)}\\ \text{Mass of oil drop,\hspace{0.17em}m}\\ \text{= Volume of the oil drop × Density of oil}\\ \text{=}\frac{4}{3}{\mathrm{\pi r}}^3\times \mathrm{\rho }\to \left(\mathrm{iii}\right)\\ \mathrm{Substituting}\mathrm{equations}\text{\hspace{0.17em}(ii)\hspace{0.17em}and\hspace{0.17em}}\left(\mathrm{iii}\right),\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\\ \mathrm{get}\\ \text{Ene=}\frac{4}{3}{\mathrm{\pi r}}^3\times \mathrm{\rho g}\\ \therefore \mathrm{r}=\sqrt[3]{\frac{3\text{Ene}}{4\mathrm{\pi \rho g}}}\\ \text{=}\sqrt[\text{3}]{\frac{{\text{3×2.55×10}}^{\text{4}}{\text{×12×1.6×10}}^{\text{-19}}}{{\text{4×3.14×1.26×10}}^{\text{3}}\text{×9.81}}}\\ \text{=}\sqrt[\text{3}]{{\text{946.09×10}}^{\text{-21}}}\\ {\text{=9.82×10}}^{\text{-7}}\text{\hspace{0.17em}m}\\ {\text{= 9.82×10}}^{\text{-4}}\text{mm}\\ \therefore \text{The radius of the oil drop is 9}.\text{82}\times \text{1}{0}^{-\text{4}}\text{mm}.\end{array}$

Q.26 Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

Ans.

(a) As the field lines must be normal to the surface of the conductor, therefore, the field lines showed in figure (a) do not represent electrostatic field lines.

(b) As the electrostatic field lines do not start from a negative charge, therefore, the field lines showed in figure (b) do not represent electrostatic field lines.

(c) Since the field lines emerge from the positive charges and repel each other, therefore, the field lines showed in figure (c) represent electrostatic field lines.

(d) As two electrostatic field lines cannot intersect each other, therefore, the field lines showed in figure (d) do not represent electrostatic field lines.

(e) The field lines showed in figure (e) do not represent electrostatic field lines because electrostatic field lines never form closed loops.

Q.27 In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ NC⁻¹ per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10⁻⁷ C-m in the negative z-direction?

Ans.

Here, dipole moment of the system, p = q × dl = −10⁻⁷ Cm

The rate of increase of electric field per unit length ,

\begin{array}{l}\frac{dE}{dl}={10}^5{\text{NC}}^{\text{-1}}\\ \text{Force experienced by the system is given as:}\\ \text{F = qE}\\ \text{=q}\frac{\text{dE}}{\text{dl}}\text{×dl}\\ \text{=p×}\frac{\text{dE}}{\text{dl}}\\ {\text{=-10}}^{\text{-7}}{\text{×10}}^{\text{5}}{\text{= -10}}^{\text{-2}}\text{N}\end{array}

\begin{array}{l}\text{The force}onthedipole\text{is}-\text{1}{0}^{-\text{2}}\text{N in the negative z}-\\ \text{direction}\\ \therefore \text{Angle between electric field and dipole moment,}\theta \\ \text{=18}0°.\\ \text{Torque is given as:}\tau \text{=pE sinθ= pE sin180°= 0}\\ \therefore \text{Torque experienced by the system is zero}.\end{array}

Q.28 (a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig.1.36(b)].
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Ans.

(a) Consider a Gaussian surface lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.

$\begin{array}{l}\text{Let charge inside the conductor}=\text{q \hspace{0.17em}}\\ \text{According to Gauss}’\text{s law},\mathrm{we}\mathrm{have}\\ \text{Flux},\mathrm{\varphi }=\overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{ds}}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}\\ \text{Here,\hspace{0.17em}}{\mathrm{\epsilon }}_0=\text{Permittivity of free space}\\ \text{As\hspace{0.17em}net\hspace{0.17em}electric\hspace{0.17em}field\hspace{0.17em}inside\hspace{0.17em}a\hspace{0.17em}charged\hspace{0.17em}conductor},\text{E}=0\\ \therefore \frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}=0\\ \mathrm{As}{\mathrm{\epsilon }}_0\ne 0\\ \therefore \mathrm{q}=0\\ \therefore \mathrm{Charge}\mathrm{inside}\mathrm{the}\mathrm{conductor}=0.\\ \therefore \mathrm{Entire}\mathrm{charge}\mathrm{Q}\mathrm{on}\mathrm{the}\mathrm{conductor}\mathrm{appears}\mathrm{on}\mathrm{the}\\ \mathrm{outer}\mathrm{surface}\mathrm{of}\mathrm{the}\mathrm{conductor}.\end{array}$

(b) Another conductor B having charge +q is inserted inside conductor A and it is insulated from A. Therefore, a charge of amount −q will be induced in the inner surface of conductor A and +q is induced on the outer surface of conductor A. Since the outer surface of conductor A has originally a charge of amount Q, therefore, total charge on the outer surface of conductor A becomes Q + q.

(c) A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it in a metallic cover.

Q.29 A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is, where is the unit vector in the outward normal direction, and is the surface charge density near the hole.

Ans.

\begin{array}{l}\text{Consider a conductor with a tiny}\text{hole}\text{into}\text{its}\text{surface}.\\ \text{Electric field inside the hole =0}\\ \text{Suppose E}is\text{electric field outside the conductor,}\text{q}is\\ \text{the}\text{electric}\text{charge}\text{and}\sigma isthesurfacech\arg edensity.\\ Electricch\arg e,q=\sigma \times ds\\ \text{According to Gauss}’\text{s law},wehave\\ {\displaystyle \oint \overrightarrow{E}}.\overrightarrow{ds}=\frac{q}{{\epsilon }_{\text{0}}}\\ Here,{\epsilon }_{\text{0}}=Permittivityoffreespace\\ \therefore Eds=\frac{\sigma \times ds}{{\epsilon }_{\text{0}}}\\ E=\frac{\sigma }{{\epsilon }_{\text{0}}}\\ or\overrightarrow{E}=\frac{\sigma }{{\epsilon }_{\text{0}}}\widehat{n}\\ \therefore \text{Electric field just outside the conductor is}\frac{\sigma }{{\epsilon }_{\text{0}}}\widehat{n}.\\ \text{This field is superposition of field due to filled}\text{up}\text{hole}\\ \text{and the field due to the rest of the}\text{charged conductor}.\\ \text{These two}\text{fields are equal and opposite inside the}\\ \text{conductor}\text{.}\text{Therefore,}\text{there}\text{is}noelectricfieldinside\\ theconductor.However,thesefieldsare\text{equal in}\\ \text{magnitude and direction outside the conductor}.\\ \therefore \text{The electric}\text{field}\text{at}P,\text{due to each}\text{part}\text{=}\frac{1}{2}\overrightarrow{E}\\ \text{=}\frac{\sigma }{2{\epsilon }_{\text{0}}}\widehat{n}\end{array}

Q.30 Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Ans.

Consider a long thin wire AB (as shown in the figure) of uniform linear charge density λ.

Take a point S at a perpendicular distance l from the mid-point O of the wire, as shown in the given figure.

\begin{array}{l}\text{Let E be the electric field at point S because of wire AB}.\\ \text{Consider a small element of length dx on the wire with}\\ \text{OZ}=\text{x}\\ \text{Let q be the charge on this element}.\\ \therefore \text{q}=\lambda dx\\ \text{Electric field due to the element},\\ dE=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dx}{{\left(SZ\right)}^2}\\ As,SZ=\sqrt{l^2+x^2}\\ \therefore dE=\frac{\lambda dx}{4\pi {\epsilon }_0\left(l^2+x^2\right)}\\ Theelectricfieldcanberesolved\mathrm{int}otworec\tan gular\\ components:\text{dE}\cos \theta perpendiculartoABand\text{dE}\sin \theta \\ paralleltoAB.\\ \text{Only the perpendicular component dEcos}\theta \text{affects}\\ \text{point S}.\\ \therefore \text{Effective component}\text{of}\text{electric field at point S due to}\\ \text{the element dx,}{\text{dE}}_{\text{1}}=\text{dE}\cos \theta \\ \therefore {\text{dE}}_{\text{1}}=\frac{\lambda \text{dx}\cos \theta }{4\pi {\epsilon }_0\left(l^2+x^2\right)}\to (i)\\ In\Delta SZO,\\ \tan \theta =\frac{x}{l}\\ \therefore x=l\tan \theta \to (ii)\\ \text{Differentiating equation}\left(ii\right),\text{we get}\\ \frac{dx}{d\theta }=l{\sec }^2\theta \\ \therefore dx=l{\sec }^2\theta d\theta \to (iii)\\ \text{From equation}\left(ii\right),\\ x^2+l^2=l^2+l^2{\tan }^2\theta \\ \therefore l^2\left(1+{\tan }^2\theta \right)=l^2{\sec }^2\theta \\ x^2+l^2=l^2{\sec }^2\theta \to (iv)\\ \text{Substituting equations}\left(iii\right)\text{and}\left(iv\right)\text{in equation}\left(i\right),\text{we}\\ get\\ d{E}_1=\frac{\lambda l{\sec }^2\theta d\theta }{4\pi {\epsilon }_0{l}^2{\sec }^2\theta }\times \cos \theta \\ \therefore d{E}_1=\frac{\lambda \cos \theta d\theta }{4\pi {\epsilon }_{0}l}\to (v)\\ \text{The wire is so long that angle}\theta \text{tends from –}\frac{\pi }{2}\text{to}+\frac{\pi }{2}.\\ \text{Integrating equation}\left(\text{5}\right),{\text{we obtain the value of E}}_{\text{1}}\text{as},\\ {\displaystyle \underset{\text{–}\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}d{E}_1}={\displaystyle \underset{\text{–}\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{\lambda \cos \theta d\theta }{4\pi {\epsilon }_{0}l}}\\ E_1=\frac{\lambda }{4\pi {\epsilon }_{0}l}{\left[\sin \theta \right]}_{\text{–}\frac{\pi }{2}}^{\frac{\pi }{2}}\\ =\frac{\lambda }{4\pi {\epsilon }_{0}l}\times 2\\ \therefore E_1=\frac{\lambda }{2\pi {\epsilon }_{0}l}\\ \therefore \text{The}\text{electric field due to long wire is}\frac{\lambda }{2\pi {\epsilon }_{0}l}.\end{array}

Q.31 It is now established that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge (+2/3) e, and the ‘down’ quark (denoted by d) of charge (−1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{t}\text{otal\hspace{0.17em}number\hspace{0.17em}of\hspace{0.17em}quarks\hspace{0.17em}in\hspace{0.17em}a\hspace{0.17em}proton=3}\\ \text{Let number\hspace{0.17em}of up quarks in a proton}=\text{n}\\ \therefore \text{Number of down quarks in a proton}=\text{3-n}\\ \text{Total\hspace{0.17em}charge\hspace{0.17em}on\hspace{0.17em}the\hspace{0.17em}proton\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ \left(\frac{2}{3}\mathrm{e}\right)\mathrm{n}+(-\frac{1}{3}\mathrm{e})\left(\text{3-n}\right)=+\mathrm{e}\\ \frac{2}{3}\mathrm{en}-\mathrm{e}+\frac{\mathrm{ne}}{3}=\mathrm{e}\\ \\ \mathrm{ne}(\frac{2}{3}+\frac{1}{3})=2\mathrm{e}\\ \therefore \mathrm{ne}=2\mathrm{e}\\ \therefore \mathrm{n}=2\\ \therefore \mathrm{Number}\mathrm{of}\mathrm{up}\mathrm{quarks}\mathrm{in}\mathrm{a}\mathrm{proton}=2\\ \mathrm{Number}\mathrm{of}\mathrm{down}\mathrm{quarks}\mathrm{in}\mathrm{a}\mathrm{proton}=3-2=1\\ \mathrm{A}\mathrm{proton}\mathrm{can}\mathrm{be}\mathrm{represented}\mathrm{as}:\mathrm{p}\Rightarrow \mathrm{UUd}\\ \text{Here,\hspace{0.17em}}\mathrm{t}\text{otal\hspace{0.17em}number\hspace{0.17em}of\hspace{0.17em}quarks\hspace{0.17em}in\hspace{0.17em}a\hspace{0.17em}proton=3}\\ \text{Let number\hspace{0.17em}of up quarks in a neutron}=\text{n}\\ \therefore \text{Number of down quarks in a neutron}=\text{3-n}\\ \text{Total\hspace{0.17em}charge\hspace{0.17em}on\hspace{0.17em}the\hspace{0.17em}neutron\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ \left(\frac{2}{3}\mathrm{e}\right)\mathrm{n}+(-\frac{1}{3}\mathrm{e})\left(\text{3-n}\right)=0\\ \frac{2}{3}\mathrm{en}-\mathrm{e}+\frac{\mathrm{ne}}{3}=\mathrm{o}\\ \mathrm{ne}(\frac{2}{3}+\frac{1}{3})=0\\ \therefore \mathrm{ne}=\mathrm{e}\\ \therefore \mathrm{n}=1\\ \therefore \mathrm{Number}\mathrm{of}\mathrm{up}\mathrm{quarks}\mathrm{in}\mathrm{a}\mathrm{neutron}=1\\ \mathrm{Number}\mathrm{of}\mathrm{down}\mathrm{quarks}\mathrm{in}\mathrm{a}\mathrm{neutron}=3-1=2\\ \mathrm{A}\mathrm{neutron}\mathrm{can}\mathrm{be}\mathrm{represented}\mathrm{as}:\mathrm{p}\Rightarrow \mathrm{Udd}\end{array}$

Q.32 (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

Ans.

(a) Suppose the equilibrium of the test charge is stable. When a test charge is displaced from its equilibrium position in any direction, it experiences a restoring force towards a null point, where the electric field is zero. It shows that there is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface not enclosing any charge is zero. Therefore, the equilibrium of the test charge cannot be stable.
(b) The mid-point of the line joining the two like charges is the null point. When a test charged is slightly displaced along the line, it experiences a restoring force. However, if it is displaced normal to the joining line, then the net force takes it further away from the null point. Therefore, the equilibrium cannot be stable.

Q.33 A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed v_x (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/ (2m v_x²).
Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.33em}}\\ \text{Charge on the particle of mass m}=-\text{q}\\ \text{Velocity of particle}={\text{v}}_{\text{x}}\\ \text{Length of plates}=\text{L}\\ \text{Magnitude of the uniform electric field}=\text{E}\\ \text{Mechanical force\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as: F}\\ =\text{Mass}\left(\text{m}\right)\times \text{Acceleration}\left(\text{a}\right)\\ \therefore \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\\ \mathrm{Electric}\mathrm{force}\mathrm{is}\mathrm{given}\mathrm{as}:\mathrm{F}=\mathrm{qE}\\ \therefore \text{Acceleration},\\ \mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}\to \left(\mathrm{i}\right)\\ \text{Time taken by the particle to cross the field is given as:}\\ \text{t=}\frac{\mathrm{distance}}{\mathrm{velocity}}\\ =\frac{\mathrm{length}\mathrm{of}\mathrm{plate}}{\mathrm{velocity}\mathrm{of}\mathrm{particle}}\\ =\frac{\mathrm{L}}{{\mathrm{v}}_{\mathrm{x}}}\to \left(\mathrm{ii}\right)\\ \text{Along vertical direction},\text{initial velocity},\\ \text{u}=0\\ \text{According to third equation of motion},\text{the vertical}\\ \text{deflection s of the particle can be\hspace{0.33em}obtained as},\\ \mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ \mathrm{s}=0+\frac{1}{2}\left(\frac{\mathrm{qE}}{\mathrm{m}}\right){\left(\frac{\mathrm{L}}{{\mathrm{v}}_{\mathrm{x}}}\right)}^2\\ \therefore \mathrm{s}=\frac{{\mathrm{qEL}}^2}{2{\mathrm{mv}}_{\mathrm{x}}^2}\to \left(\mathrm{iii}\right)\\ \therefore \text{Vertical deflection of the particle at the far edge of}\\ \text{the plate is\hspace{0.33em}}\frac{{\mathrm{qEL}}^2}{2{\mathrm{mv}}_{\mathrm{x}}^2}.\\ \text{This is exacly\hspace{0.33em}similar to the motion of horizontal}\\ \text{projectiles under the\hspace{0.33em}effect\hspace{0.33em}of\hspace{0.33em}gravity}.\end{array}$

Q.34 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity v_x= 2.0 × 10⁶ m s⁻¹. If E between the plates separated by 0.5 cm is 9.1 × 10² NC^-1, where will the electron strike the upper plate? (| e | =1.6 × 10⁻¹⁹ C, m_e = 9.1 × 10⁻³¹ kg.)

Ans.

\begin{array}{l}\text{Here,}{\text{velocity of particle, v}}_{\text{x}}\text{= 2}{\text{.0×10}}^{\text{6}}{\text{ms}}^{\text{-1}}\\ \text{Separation of the plates, d = 0}\text{.5 cm = 0}\text{.005 m}\\ \text{Electric field between the plates, E=9}{\text{.1×10}}^{\text{2}}{\text{NC}}^{\text{-1}}\\ \text{Charge of electron, q = 1}{\text{.6 × 10}}^{\text{-19}}\text{C}\\ {\text{Mass of electron, m}}_{\text{e}}\text{= 9}{\text{.1 × 10}}^{\text{-31}}\text{kg}\\ \text{Let the electron strikes the upper plate at the end of}\\ \text{plate L,}\text{as}\text{soon}\text{as}\text{its deflection is d}\text{.}\\ \therefore d=\frac{{\text{qEL}}^2}{2m{v}_x{}^2}\\ \therefore L=\sqrt{\frac{2\times 0.005\times 9.1\times {10}^{-31}\times {\left(\text{2}.0\times \text{1}{0}^{\text{6}}\right)}^2}{\text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\times \text{9}.\text{1}\times \text{1}{0}^{\text{2}}}}\\ =\sqrt{0.025\times {10}^{-2}}\\ =\sqrt{2.5\times {10}^{-4}}\\ =1.6\times {10}^{-2}m\\ =1.6cm\\ \therefore \text{The}\text{electron will strike the upper plate after}\\ \text{travelling 1}.\text{6 cm}.\end{array}