Electric Charges is the first electrostatics chapter in Class 12 Physics. Electric Charges and Fields Class 12 Important Questions cover electric charge, Coulomb’s law, superposition, electric field, electric field lines, dipole moment, electric flux, Gauss’s law, numericals, and 5-mark derivations.
A charged comb attracting paper, lightning during a storm, and the force between two tiny charged particles all come from the same idea: electric charges create fields around them.
Physics Class 12 Chapter 1 Important Questions
Q.
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?
Q.
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?
Q.
A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vx (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 vx2).
Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.
Q.
(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.
Q.
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.
Q.
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.]
Q.
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.
Q.
(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.
Q.
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?
Q.
Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?
Q.
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).
Q.
Q.
An infinite line charge produces a field of 9 × 104 NC-1 at a distance of 2 cm. Calculate the linear charge density.
Q.
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?
Q.
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?
Q.
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?
Q.
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.)
Q.
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?
Q.
A system has two charges qA = 2.5 × 10−7 C and qB = −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?
Q.
Check that the ratio ke2/Gmemp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Q.
(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?
Q.
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.
Q.
Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, and qD = −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?
Q.
(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?
Q.
Two point charges qA = 3 μC and qB = −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?
Q.
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.
Q.
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?
Q.
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?
Q.
(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?
Q.
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?
Q.
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?
Q.
Q.
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?
Q.
Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx= 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, me = 9.1 × 10−31 kg.)
Class 12 Physics Chapter 1 becomes easier when students separate three things clearly: force between charges, field due to charges, and flux through a surface. Once these basics are clear, Coulomb’s law numericals, dipole questions, and Gauss’s law derivations become much easier to solve.
Key Takeaways
| Topic |
What to Focus On |
| Electric Charge |
Additivity, conservation, quantisation |
| Coulomb’s Law |
Formula, direction, vector form, medium effect |
| Superposition Principle |
Net force and net field due to multiple charges |
| Electric Field |
Field due to point charge and system of charges |
| Electric Field Lines |
Direction, properties, non-intersection |
| Electric Dipole |
Dipole moment, torque, axial and equatorial fields |
| Electric Flux |
Formula, angle dependence, SI unit |
| Gauss’s Law |
Statement and applications |
| Numericals |
Coulomb force, flux, dipole torque, field at a point |
| Exam Pattern |
MCQ, assertion-reason, 2-mark, 3-mark, 5-mark questions |
Class 12 Physics Chapter List
Important Topics in Electric Charges and Fields for 2026 Exams
Class 12 physics chapter 1 important questions mostly come from formulas, field direction, and Gauss’s law applications.
Before solving numericals, revise the meaning of charge, field, flux, and dipole moment. These decide which formula applies.
- Electric charge and its properties
- Quantisation and conservation of charge
- Coulomb’s law in vacuum and medium
- Vector form of Coulomb’s law
- Principle of superposition
- Electric field due to a point charge
- Electric field due to a system of charges
- Electric field lines and their properties
- Electric dipole and dipole moment
- Torque on a dipole in a uniform electric field
- Electric flux and its SI unit
- Gauss’s law
- Field due to infinite wire, sheet, and spherical shell
- Coulomb’s law, flux, and dipole numericals
Important Questions Class 12 Physics Chapter 1 with Answers
Electric charges and fields important questions should be practised in order: definitions, formula-based questions, numericals, assertion-reason, and derivations.
These class 12 physics chapter 1 important questions with answers cover the main concepts asked in school exams and board-style papers.
Very Short Answer Questions from Electric Charges and Fields
One-mark questions test units, definitions, and direct properties. Keep answers short and exact.
Q1. Name the physical quantity whose SI unit is Cm.
Ans. Electric dipole moment.
Its SI unit is coulomb metre.
Q2. Name the physical quantity whose SI unit is NC⁻¹.
Ans. Electric field intensity.
It is a vector quantity.
Q3. What is 1 coulomb?
Ans. A body carries a charge of 1 coulomb if it repels an equal charge placed 1 m away in vacuum with a force of 9 × 10⁹ N.
Q4. Does the force between two point charges change if the dielectric constant of the medium increases?
Ans. Yes.
Force decreases when dielectric constant increases.
Fmedium = Fvacuum / K
Q5. What is the electric flux through a cube if charge q is placed at its centre?
Ans. Total flux through the cube is q/ε₀.
Flux through each face is q/6ε₀.
Q6. What does “electric charge is quantised” mean?
Ans. It means charge on a body is always an integral multiple of elementary charge.
q = ne
Here, n is an integer and e = 1.6 × 10⁻¹⁹ C.
Q7. A charge q is placed at the centre of a sphere. What happens to flux if the radius is doubled?
Ans. Flux remains unchanged.
By Gauss’s law, flux depends only on enclosed charge, not on radius.
Q8. Which physical quantity has SI unit JC⁻¹? Is it scalar or vector?
Ans. Electric potential has SI unit JC⁻¹.
It is a scalar quantity.
Short Answer Questions on Electric Charge and Coulomb’s Law
Coulomb’s law questions test both magnitude and direction. Always check whether the force is attractive or repulsive.
Physics Class 12 Chapter 1 Questions and Answers
Q1. Two charges 2 × 10⁻⁷ C and 3 × 10⁻⁷ C are placed 30 cm apart in air. Find the force between them.
Ans.
F = kq₁q₂/r²
q₁ = 2 × 10⁻⁷ C
q₂ = 3 × 10⁻⁷ C
r = 30 cm = 0.3 m
F = 9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷ / (0.3)²
F = 9 × 10⁹ × 6 × 10⁻¹⁴ / 0.09
F = 6 × 10⁻³ N
The force is repulsive because both charges are positive.
Q2. Why can two electric field lines never intersect each other?
Ans. If two electric field lines intersect, the point of intersection will have two tangents.
That means the electric field will have two directions at the same point.
This is impossible because the electric field at a point has a unique direction.
Q3. Why is quantisation of electric charge not important at macroscopic scale?
Ans. At macroscopic scale, charges are very large compared to elementary charge.
The small value of e = 1.6 × 10⁻¹⁹ C becomes negligible.
So, charge appears continuous at large scale.
Q4. When a glass rod is rubbed with silk, charges appear on both. How does this agree with conservation of charge?
Ans. Rubbing transfers electrons from the glass rod to silk.
The glass rod becomes positively charged.
The silk becomes negatively charged.
Equal and opposite charges are produced, so total charge remains conserved.
Q5. An electric dipole is held at 30° to a uniform electric field of 10⁴ N/C and experiences torque of 9 × 10⁻²⁶ Nm. Find the dipole moment.
Ans.
τ = pE sinθ
9 × 10⁻²⁶ = p × 10⁴ × sin30°
9 × 10⁻²⁶ = p × 10⁴ × 0.5
p = 9 × 10⁻²⁶ / 5 × 10³
p = 1.8 × 10⁻²⁹ Cm
Q6. Why is an electrostatic field line a continuous curve with no sudden breaks?
Ans. A charge experiences continuous force in an electrostatic field.
It cannot jump from one point to another.
So, the field line representing its path must be continuous.
Q7. What is electric dipole moment? Give its SI unit.
Ans. Electric dipole moment is the product of the magnitude of either charge and the separation between the charges.
p = q × 2a
It is directed from negative charge to positive charge.
SI unit = coulomb metre.
Electric Field and Electric Field Lines Questions
Electric field questions test force per unit charge and direction.
For field line questions, remember that field lines start from positive charges and end at negative charges.
Chapter 1 Physics Class 12 Important Questions
Q1. Define electric field at a point. Give its SI unit.
Ans. Electric field at a point is the force experienced by a unit positive test charge placed at that point.
E = F/q
SI unit = NC⁻¹ or Vm⁻¹
Q2. State four properties of electric field lines.
Ans.
- Electric field lines start from positive charges and end at negative charges.
- Two field lines never intersect.
- Field lines are continuous curves in a charge-free region.
- Electrostatic field lines do not form closed loops.
Q3. For a positive point charge, in which direction does the electric field point?
Ans. For a positive point charge, electric field points radially outward.
For a negative point charge, electric field points radially inward.
Q4. Two point charges qA = 3 μC and qB = -3 μC are placed 20 cm apart. Find the electric field at the midpoint.
Ans.
Distance of midpoint from each charge:
r = 10 cm = 0.1 m
Field due to qA:
E = kq/r²
E = 9 × 10⁹ × 3 × 10⁻⁶ / (0.1)²
E = 2.7 × 10⁶ NC⁻¹
Direction is from A to B.
Field due to qB also points from A to B because qB is negative.
Total field:
E = 2.7 × 10⁶ + 2.7 × 10⁶
E = 5.4 × 10⁶ NC⁻¹
Direction is from A to B.
Electric Charges and Fields Important Numerical Questions
Class 12 physics chapter 1 numericals are high-scoring when formulas and units are written clearly.
Use SI units before substitution. Convert cm to metre and μC to C first.
Numericals on Coulomb’s Law and Force Between Charges
Q1. Two small spheres carry charges 2 × 10⁻⁷ C and 3 × 10⁻⁷ C. They are placed 30 cm apart in air. Find the electrostatic force between them.
Ans.
F = kq₁q₂/r²
F = 9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷ / (0.3)²
F = 6 × 10⁻³ N
The force is repulsive.
Q2. Two insulated copper spheres A and B have centres 50 cm apart. Each carries charge 6.5 × 10⁻⁷ C. Find the mutual force. What happens if each charge is doubled and distance is halved?
Ans.
For original charges:
F = kq²/r²
F = 9 × 10⁹ × (6.5 × 10⁻⁷)² / (0.5)²
F = 1.52 × 10⁻² N
If each charge is doubled, force becomes 4 times.
If distance is halved, force becomes 4 times.
Total change = 4 × 4 = 16
New force:
F’ = 16 × 1.52 × 10⁻²
F’ = 0.243 N
Q3. A polythene piece rubbed with wool has charge -3 × 10⁻⁷ C. How many electrons were transferred? Is there transfer of mass?
Ans.
n = q/e
n = 3 × 10⁻⁷ / 1.6 × 10⁻¹⁹
n = 1.875 × 10¹² electrons
Electrons were transferred from wool to polythene.
Mass transferred:
m = n × me
m = 1.875 × 10¹² × 9.1 × 10⁻³¹
m = 1.706 × 10⁻¹⁸ kg
This mass is negligibly small.
Numericals on Electric Field, Dipole Moment and Flux
Q1. A system has charges qA = 2.5 × 10⁻⁷ C at A(0, 0, -15 cm) and qB = -2.5 × 10⁻⁷ C at B(0, 0, +15 cm). Find total charge and dipole moment.
Ans.
Total charge:
qA + qB = 2.5 × 10⁻⁷ - 2.5 × 10⁻⁷
= 0
Separation = 30 cm = 0.3 m
Dipole moment:
p = q × d
p = 2.5 × 10⁻⁷ × 0.3
p = 7.5 × 10⁻⁸ Cm
Direction is along positive z-axis.
Q2. Uniform electric field E = 3 × 10³ î N/C exists. Find flux through a square of side 10 cm when the plane is parallel to yz-plane and when its normal makes 60° with x-axis.
Ans.
Area:
A = 0.1 × 0.1
A = 0.01 m²
Case 1:
Plane is parallel to yz-plane, so normal is along x-axis.
θ = 0°
φ = EA cosθ
φ = 3 × 10³ × 0.01 × cos0°
φ = 30 Nm²/C
Case 2:
θ = 60°
φ = EA cos60°
φ = 3 × 10³ × 0.01 × 0.5
φ = 15 Nm²/C
Q3. A point charge +10 μC is 5 cm above the centre of a square of side 10 cm. Find electric flux through the square.
Ans.
Imagine the square as one face of a cube of edge 10 cm.
The charge is at the centre of the cube.
Total flux through cube:
φ = q/ε₀
φ = 10 × 10⁻⁶ / 8.854 × 10⁻¹²
φ = 1.13 × 10⁶ Nm²/C
Flux through one face:
φface = 1.13 × 10⁶ / 6
φface = 1.88 × 10⁵ Nm²/C
Q4. A dipole has dipole moment 10⁻⁷ Cm in negative z-direction. Electric field increases uniformly in positive z-direction at 10⁵ NC⁻¹ per metre. Find force and torque.
Ans.
Force on dipole:
F = p(dE/dl)
F = -10⁻⁷ × 10⁵
F = -10⁻² N
So, force is in negative z-direction.
Torque:
τ = pE sin180°
τ = 0
The dipole is antiparallel to the field direction, so torque is zero.
Assertion-Reason and MCQs from Electric Charges and Fields
Assertion-reason and MCQs test conceptual precision.
Read directions, signs, and field-line statements carefully.
Assertion-Reason Questions
Q1. Assertion (A): Two electric field lines can never intersect. Reason (R): At any point in space, electric field has a unique direction.
Ans. Both A and R are true, and R is the correct explanation of A.
If field lines intersect, the field at that point will have two directions. This is impossible.
Q2. Assertion (A): Inside a uniformly charged thin spherical shell, the electric field is zero. Reason (R): Charges on the shell produce equal and opposite fields at every interior point.
Ans. Both A and R are true, and R is the correct explanation of A.
By Gauss’s law, electric field inside a uniformly charged spherical shell is zero.
Q3. Assertion (A): Net electric flux through a closed surface is zero if no charge is enclosed. Reason (R): Electric flux depends only on charge inside the Gaussian surface.
Ans. Both A and R are true, and R is the correct explanation of A.
External charges may create field through the surface, but their net flux through a closed surface is zero.
MCQ Questions
Q1. A charge q is at the centre of a sphere of radius r. If radius becomes 2r, electric flux through the sphere is:
(a) Doubled
(b) Halved
(c) Same
(d) Quadrupled
Ans. (c) Same
Flux = q/ε₀, independent of surface size.
Q2. The SI unit of electric dipole moment is:
(a) NC⁻¹
(b) Cm
(c) Nm
(d) JC⁻¹
Ans. (b) Cm
Q3. Two equal and opposite charges +q and -q are placed 2a apart. Electric field on the equatorial line at distance r, where r >> a, varies as:
(a) 1/r²
(b) 1/r
(c) 1/r³
(d) 1/r⁴
Ans. (c) 1/r³
Dipole field varies as 1/r³ at large distances.
Long Answer Questions and Derivations from Electric Charges and Fields
Five-mark questions usually test derivations and multi-step reasoning.
Write the diagram, formula, direction, and final result clearly.
Chapter 1 Physics Class 12 Important Numerical Questions
Q1. Four charges qA = 2 μC, qB = -5 μC, qC = 2 μC, and qD = -5 μC are placed at the corners of a square of side 10 cm. Find the force on a 1 μC charge at the centre.
Ans.
The distance from each corner to the centre is equal.
Charges qA and qC are equal and placed at opposite corners.
The forces due to qA and qC on the centre charge are equal and opposite.
So, they cancel.
Charges qB and qD are also equal and placed at opposite corners.
Their forces are also equal and opposite.
So, they cancel.
Net force on the 1 μC charge at the centre is zero.
Q2. Check whether ke²/(Gmemp) is dimensionless. Find its value and state its significance.
Ans.
Unit of ke²:
k has unit Nm²C⁻².
e² has unit C².
So, ke² has unit Nm².
Unit of Gmemp:
G has unit Nm²kg⁻².
me and mp have unit kg each.
So, Gmemp also has unit Nm².
Therefore, the ratio is dimensionless.
Using values:
k = 9 × 10⁹ Nm²C⁻²
e = 1.6 × 10⁻¹⁹ C
G = 6.67 × 10⁻¹¹ Nm²kg⁻²
me = 9.1 × 10⁻³¹ kg
mp = 1.67 × 10⁻²⁷ kg
ke²/(Gmemp) ≈ 2.3 × 10³⁹
This means electrostatic force between a proton and electron is about 2.3 × 10³⁹ times stronger than gravitational force between them.
Q3. Derive the expression for electric field at a point on the axial line of a short electric dipole.
Ans.
Consider an electric dipole with charges +q and -q separated by distance 2a.
Let O be the centre of the dipole.
Let point P lie on the axial line at distance r from O.
Distance of P from +q:
r - a
Distance of P from -q:
r + a
Field at P due to +q:
E+ = kq/(r - a)²
Direction is along the axis away from +q.
Field at P due to -q:
E- = kq/(r + a)²
Direction is towards -q.
Since E+ > E-, net field is:
E = E+ - E-
E = kq[1/(r - a)² - 1/(r + a)²]
E = kq[(r + a)² - (r - a)²] / (r² - a²)²
E = kq[4ar] / (r² - a²)²
Since p = q × 2a,
E = 2kpr / (r² - a²)²
For a short dipole, r >> a.
So, r² - a² ≈ r²
E = 2kp/r³
Since k = 1/4πε₀,
E = 2p / 4πε₀r³
Direction is along the dipole moment.
Extra Questions with Answers from Electric Charges and Fields
These extra questions target areas where students often lose marks in board-style answers.
Use them after completing NCERT examples.
Q1. A conducting sphere of radius 10 cm has an unknown charge. The electric field 20 cm from the centre is 1.5 × 10³ NC⁻¹ directed inward. Find the net charge.
Ans.
For points outside a charged conducting sphere:
E = kq/r²
r = 20 cm = 0.2 m
1.5 × 10³ = 9 × 10⁹ × |q| / (0.2)²
|q| = 1.5 × 10³ × 0.04 / 9 × 10⁹
|q| = 6.67 × 10⁻⁹ C
Since field is directed inward, charge is negative.
q = -6.67 nC
Q2. A small test charge is placed at the null point of an electrostatic field configuration. Show that the equilibrium is unstable.
Ans.
Assume the equilibrium is stable.
Then a small displacement should bring the test charge back to the null point.
This means field lines around the null point must point inward from all directions.
Now take a small closed surface around the null point.
If all field lines point inward, net flux through the surface is negative.
But no charge is enclosed at the null point.
By Gauss’s law, net flux must be zero.
This contradiction shows that stable equilibrium is impossible.
So, the equilibrium is necessarily unstable.
Q3. A net outward flux of 8 × 10³ Nm²/C passes through a closed box. What is the net charge inside?
Ans.
Using Gauss’s law:
φ = q/ε₀
q = φε₀
q = 8 × 10³ × 8.854 × 10⁻¹²
q = 7.08 × 10⁻⁸ C
q ≈ 0.07 μC
Most Important Questions from Class 12 Physics Chapter 1 for 2026 Exams
These physics chapter 1 class 12 important questions are high-priority for revision.
Practise derivations with diagrams and numericals with full SI-unit conversion.
- Derive electric field on the axial line of an electric dipole.
- State and prove Gauss’s law.
- Use Gauss’s law to find electric field due to an infinitely long charged wire.
- State the superposition principle and solve a multi-charge force question.
- Find number of electrons transferred when charge is 3 × 10⁻⁷ C.
- Explain why two electric field lines never intersect.
- Define electric flux and write its SI unit.
- Find flux through a closed surface enclosing charge q.
- Derive torque on an electric dipole in a uniform electric field.
- Use Gauss’s law to find field inside and outside a charged spherical shell.
Important Formulas from Electric Charges and Fields
| Formula |
Expression |
| Coulomb’s law |
F = kq₁q₂/r² = q₁q₂/4πε₀r² |
| Electric field due to point charge |
E = kq/r² |
| Electric field |
E = F/q |
| Dipole moment |
p = q × 2a |
| Axial field of short dipole |
E = 2p/4πε₀r³ |
| Equatorial field of short dipole |
E = p/4πε₀r³ |
| Torque on dipole |
τ = pE sinθ |
| Electric flux |
φ = E · A = EA cosθ |
| Gauss’s law |
φ = q/ε₀ |
| Field due to infinite wire |
E = λ/2πε₀r |
| Field due to infinite sheet |
E = σ/2ε₀ |
| Field outside spherical shell |
E = q/4πε₀r² |
| Field inside spherical shell |
E = 0 |
Common Mistakes in Electric Charges and Fields Questions
Most mistakes in this chapter happen because students treat vector quantities like scalars.
Draw the charge diagram before solving any numerical.
Force Direction Mistakes
Coulomb’s law gives magnitude.
Direction depends on charge signs.
Like charges repel. Unlike charges attract.
Vector Addition Mistakes
In multi-charge systems, forces and fields are vectors.
Do not add magnitudes directly unless all vectors act along the same line and direction.
Use components when needed.
Dipole Formula Mistakes
On the axial line:
E = 2p/4πε₀r³
On the equatorial line:
E = p/4πε₀r³
The equatorial field is opposite to dipole moment.
Flux Mistakes
Electric flux through a closed surface depends only on enclosed charge.
External charges do not change net flux through a closed surface.
Field Line Mistakes
Electric field lines start from positive charges and end on negative charges.
They never intersect.
They do not form closed loops in electrostatics.
Unit Mistakes
Electric field unit is NC⁻¹.
Electric flux unit is Nm²/C.
Dipole moment unit is Cm.
Do not mix these units.
Marks Distribution for Class 12 Physics Chapter 1
| Question Type |
Topics Usually Tested |
| 1 mark |
Units, charge properties, flux, field direction |
| 2 marks |
Field lines, dipole moment, short numericals |
| 3 marks |
Coulomb’s law, flux, dipole torque, superposition |
| 5 marks |
Gauss’s law, dipole field derivation, spherical shell |
| MCQs |
Electric field lines, flux, dipole moment, Gauss’s law |
| Assertion-reason |
Field lines, shell field, Gauss’s law, quantisation |