Important Questions Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance

Electrostatic potential is work done per unit positive charge in moving a charge between two points. Capacitance measures how much charge a conductor system can store for a given potential difference.

Electric potential turns force-based electrostatics into an energy-based method for solving charge systems and capacitor circuits. Important Questions Class 12 Physics Chapter 2 help students practise potential energy, potential due to charges, dipoles, equipotential surfaces, conductors, dielectrics, capacitance, capacitor combinations, and stored energy. The CBSE 2026 chapter contains formula-based numericals, conceptual conductor properties, dielectric logic, and NCERT capacitor exercises.

Key Takeaways

• Potential Difference: It equals work done per unit positive test charge between two points.
• Point Charge Potential: Potential due to charge Q is V = Q/(4πε0r).
• Conductor Interior: Electrostatic field inside a conductor is zero in static conditions.
• Capacitor Energy: Stored energy equals U = 1/2 CV².

Important Questions Class 12 Physics Chapter 2 Structure 2026

Important Questions Class 12 Physics Chapter 2 with Answers

Potential and capacitance questions connect work, charge, field, and energy.
Students should identify whether the question uses potential, capacitance, conductor properties, or energy.
These electrostatic potential and capacitance class 12 important questions follow the NCERT 2026 sequence.

1. What does Important Questions Class 12 Physics Chapter 2 mainly test?

Important Questions Class 12 Physics Chapter 2 mainly test potential, potential energy, equipotential surfaces, conductors, dielectrics, capacitance, and capacitor energy. The chapter combines definitions, derivations, and numerical problems.

1. Potential Skill: Use V = W/q.
2. Charge System Skill: Add potentials algebraically.
3. Capacitor Skill: Use C = Q/V.
4. Energy Skill: Use U = 1/2 CV².
5. Final Result: The chapter tests electrostatic energy and charge storage.

2. Why is electrostatic force called conservative?

Electrostatic force is conservative because work depends only on initial and final points. It does not depend on the path.

1. Charge Movement: A charge moves between two points.
2. Work Done: Electrostatic work depends on endpoints.
3. Path Role: Path shape does not change the work.
4. Final Result: Electrostatic force is path independent.

3. Why is potential difference more important than absolute potential?

Potential difference is more important because only differences affect work and energy. Absolute potential can shift by an arbitrary constant.

1. Work Relation: W = q(VP − VR).
2. Zero Choice: Potential at infinity is often taken as zero.
3. Physical Meaning: Only VP − VR affects charge movement.
4. Final Result: Potential difference is physically significant.

Electric Potential Energy Class 12 Questions

Electric potential energy comes from work done against electrostatic force.
It becomes positive for like charges and negative for unlike charges.
These electric potential energy class 12 questions cover work, charge systems, and external fields.

4. What is electric potential energy in electrostatics?

Electric potential energy is stored energy due to the position of charges in an electric field. It equals external work done against electrostatic force.

1. External Work: Charge moves slowly without acceleration.
2. Energy Storage: Work appears as potential energy.
3. Conservative Field: Final value depends on positions.
4. Final Result: Electric potential energy equals stored electrostatic work.

5. What is potential energy of two point charges?

The potential energy of two point charges is U = q1q2/(4πε0r). Here r is their separation.

1. Charge 1: q1.
2. Charge 2: q2.
3. Separation: r.
4. Formula Used: U = q1q2/(4πε0r).
5. Final Result: Potential energy depends on charge product and separation.

6. When is potential energy of two charges positive?

Potential energy is positive when both charges have the same sign. External work is needed to bring like charges together.

1. Condition: q1q2 > 0.
2. Force Nature: Repulsive.
3. Work Required: Positive external work.
4. Final Result: Like charges have positive potential energy.

7. When is potential energy of two charges negative?

Potential energy is negative when charges have opposite signs. Attraction helps the charges come together from infinity.

1. Condition: q1q2 < 0.
2. Force Nature: Attractive.
3. Work From Infinity: Negative external work.
4. Final Result: Unlike charges have negative potential energy.

8. Two charges 7 μC and −2 μC are 18 cm apart. Find potential energy.

The potential energy is −0.7 J. Use U = q1q2/(4πε0r).

1. Given Data:
   q1 = 7 × 10^-6 C
   q2 = −2 × 10^-6 C
   r = 0.18 m
   1/(4πε0) = 9 × 10^9
2. Formula Used: U = q1q2/(4πε0r).
3. Calculation:
   U = 9 × 10^9 × 7 × 10^-6 × (−2 × 10^-6)/0.18
   U = −0.7 J
4. Final Result: U = −0.7 J.

Electrostatic Potential Class 12 Physics Questions

Electrostatic potential converts work done into a value per unit charge.
It lets students solve charge configurations without handling vector force directions.
These electrostatic potential class 12 physics questions cover definition, unit, and signs.

9. What is electrostatic potential in Class 12 Physics?

Electrostatic potential is work done per unit positive test charge in bringing it from infinity to a point. Its SI unit is volt.

1. Work Done: W.
2. Test Charge: q.
3. Formula Used: V = W/q.
4. SI Unit: Volt.
5. Final Result: Electrostatic potential equals work per unit charge.

10. What is potential difference between two points?

Potential difference is work done per unit charge in moving a positive test charge between two points. It equals VP − VR.

1. Initial Point: R.
2. Final Point: P.
3. Formula Used: VP − VR = W/q.
4. Final Result: Potential difference measures work per unit charge.

11. Why is electrostatic potential a scalar quantity?

Electrostatic potential is scalar because it comes from work per unit charge. Potentials add algebraically, not vectorially.

1. Quantity Type: Work is scalar.
2. Charge: Scalar.
3. Ratio: V = W/q.
4. Final Result: Potential is a scalar quantity.

12. What is one electron volt?

One electron volt is the energy gained by an electron across 1 volt. Its value is 1.6 × 10^-19 J.

1. Charge Magnitude: e = 1.6 × 10^-19 C.
2. Potential Difference: 1 V.
3. Energy: E = qV.
4. Final Result: 1 eV = 1.6 × 10^-19 J.

Potential Due to Point Charge Class 12 Questions

A point charge creates potential that varies inversely with distance.
Unlike electric field, potential does not require vector components.
These potential due to point charge class 12 questions cover formula, signs, and numericals.

13. What is potential due to a point charge?

Potential due to a point charge is V = Q/(4πε0r). It is measured from infinity as zero potential.

1. Source Charge: Q.
2. Distance: r.
3. Formula Used: V = Q/(4πε0r).
4. Final Result: Potential varies as 1/r.

14. What is the sign of potential due to positive and negative charges?

Potential is positive for a positive charge and negative for a negative charge. The sign follows the sign of Q.

1. Formula: V = Q/(4πε0r).
2. Positive Q: V > 0.
3. Negative Q: V < 0.
4. Final Result: Potential has the same sign as source charge.

15. How does potential vary with distance from a point charge?

Potential decreases as 1/r with distance. It becomes zero at infinity.

1. Formula Used: V ∝ 1/r.
2. Distance Doubled: Potential becomes half.
3. Infinity: V = 0.
4. Final Result: Point charge potential falls as 1/r.

16. Find potential at 9 cm due to charge 4 × 10^-7 C.

The potential is 4 × 10^4 V. Use V = Q/(4πε0r).

1. Given Data:
   Q = 4 × 10^-7 C
   r = 9 cm = 0.09 m
   1/(4πε0) = 9 × 10^9
2. Formula Used: V = Q/(4πε0r).
3. Calculation:
   V = 9 × 10^9 × 4 × 10^-7/0.09
   V = 4 × 10^4 V
4. Final Result: V = 4 × 10^4 V.

17. Find work done in bringing 2 × 10^-9 C to that point.

The work done is 8 × 10^-5 J. Use W = qV.

1. Given Data:
   q = 2 × 10^-9 C
   V = 4 × 10^4 V
2. Formula Used: W = qV.
3. Calculation:
   W = 2 × 10^-9 × 4 × 10^4
   W = 8 × 10^-5 J
4. Final Result: W = 8 × 10^-5 J.

Electric Dipole Potential Class 12 Questions

Dipole potential depends on distance and angle from the dipole axis.
It falls faster than potential due to a single point charge.
These electric dipole potential class 12 questions cover axial, equatorial, and angular cases.

18. What is potential due to an electric dipole?

Potential due to an electric dipole is V = p cos θ/(4πε0r²) for r much larger than dipole size.

1. Dipole Moment: p.
2. Distance: r.
3. Angle: θ between p and r.
4. Formula Used: V = p cos θ/(4πε0r²).
5. Final Result: Dipole potential depends on r and θ.

19. What is potential on the axial line of an electric dipole?

Potential on the axial line is V = ±p/(4πε0r²). The sign depends on the side of the dipole.

1. Axial Direction: θ = 0° or 180°.
2. Formula Used: V = p cos θ/(4πε0r²).
3. At θ = 0°: V is positive.
4. At θ = 180°: V is negative.
5. Final Result: Axial dipole potential is ±p/(4πε0r²).

20. What is potential on the equatorial plane of an electric dipole?

Potential on the equatorial plane is zero. The two charge potentials cancel.

1. Equatorial Angle: θ = 90°.
2. Formula Used: V = p cos θ/(4πε0r²).
3. Calculation: cos 90° = 0.
4. Final Result: Equatorial dipole potential is zero.

21. How does dipole potential differ from point charge potential?

Dipole potential falls as 1/r², while point charge potential falls as 1/r. Dipole potential also depends on angle.

1. Point Charge: V ∝ 1/r.
2. Dipole: V ∝ 1/r².
3. Angle Factor: Dipole has cos θ.
4. Final Result: Dipole potential has angular dependence.

Equipotential Surfaces Class 12 Questions

Equipotential surfaces contain points at the same electric potential.
No work is needed to move a charge along such a surface.
These equipotential surfaces class 12 questions cover surface shapes and field direction.

22. What is an equipotential surface?

An equipotential surface is a surface whose every point has the same potential. Moving charge along it needs no work.

1. Potential: Constant over the surface.
2. Potential Difference: Zero between any two points.
3. Work: W = qΔV = 0.
4. Final Result: No work is done on an equipotential surface.

23. What are equipotential surfaces for a point charge?

Equipotential surfaces for a point charge are concentric spheres. The charge lies at the common centre.

1. Formula: V = Q/(4πε0r).
2. Constant Potential: r must remain constant.
3. Surface Shape: Sphere.
4. Final Result: Point charge equipotentials are concentric spheres.

24. What are equipotential surfaces in a uniform electric field?

Equipotential surfaces in a uniform electric field are planes perpendicular to the field. Potential changes along the field direction.

1. Field Direction: Along one straight direction.
2. Constant Potential Planes: Normal to field.
3. Example: If E is along x, equipotentials are y-z planes.
4. Final Result: Uniform field has plane equipotential surfaces.

25. Why is electric field perpendicular to equipotential surface?

Electric field is perpendicular because any tangential component would do work along the surface. That would change potential.

1. Equipotential Surface: ΔV = 0.
2. Work Relation: W = qΔV.
3. Tangential Field: Would make W non-zero.
4. Final Result: Electric field is normal to equipotential surfaces.

26. What is the relation between electric field and potential?

Electric field equals the negative potential gradient. In one dimension, E = −dV/dl.

1. Direction: Field points towards decreasing potential.
2. Magnitude: Rate of potential decrease per distance.
3. Formula Used: E = −dV/dl.
4. Final Result: Electric field points along steepest potential decrease.

Electrostatics of Conductors Class 12 Questions

Conductors contain mobile charges that rearrange in electrostatic equilibrium.
This rearrangement makes the field inside a conductor zero.
These electrostatics of conductors class 12 questions cover field, charge, potential, and shielding.

27. What is electric field inside a conductor in electrostatic equilibrium?

Electric field inside a conductor is zero in electrostatic equilibrium. Free charges move until the internal field cancels.

1. Conductor Property: It has mobile charges.
2. If E Existed: Charges would drift.
3. Static Condition: No current flows.
4. Final Result: E = 0 inside a conductor.

28. Where does excess charge reside on a conductor?

Excess charge resides only on the outer surface of a conductor. The conductor interior has no net excess charge.

1. Inside Field: E = 0.
2. Gauss Law: Enclosed charge inside is zero.
3. Charge Location: Surface.
4. Final Result: Excess charge lies on the conductor surface.

29. What is potential inside a conductor?

Potential is constant throughout a conductor and on its surface. No work is needed inside it.

1. Inside Field: E = 0.
2. Potential Difference: ΔV = 0.
3. Result: Same potential everywhere inside.
4. Final Result: A conductor is an equipotential body.

30. What is electric field just outside a charged conductor?

Electric field just outside a charged conductor is E = σ/ε0 normal to the surface. It points outward for positive charge.

1. Surface Charge Density: σ.
2. Formula Used: E = σ/ε0.
3. Direction: Normal to the surface.
4. Final Result: Surface field equals σ/ε0.

31. What is electrostatic shielding?

Electrostatic shielding is the protection of a region from external electric fields by a conductor. A cavity inside a conductor has zero field.

1. Conductor Condition: E = 0 inside metal.
2. Charge-Free Cavity: Field remains zero.
3. Use: Sensitive instruments use shielding.
4. Final Result: A conductor shields its cavity from external fields.

Dielectrics and Polarisation Class 12 Questions

Dielectrics cannot conduct charge freely, but their molecules can polarise.
An external electric field induces or aligns molecular dipoles inside them.
These dielectric constant class 12 questions also cover polarisation and field reduction.

32. What is a dielectric in Class 12 Physics?

A dielectric is a non-conducting substance with negligible free charge carriers. It polarises in an external electric field.

1. Charge Carriers: Very few free carriers.
2. External Field: Distorts or aligns molecules.
3. Effect: Net dipole moment appears.
4. Final Result: Dielectrics polarise but do not conduct freely.

33. What is polarisation in a dielectric?

Polarisation is dipole moment per unit volume of a dielectric. Its symbol is P.

1. Dipole Moment: Net molecular dipole moment.
2. Volume: Volume of dielectric sample.
3. Formula Meaning: P = dipole moment/volume.
4. Final Result: Polarisation measures dipole density.

34. What is the relation between polarisation and electric field?

For a linear isotropic dielectric, P = ε0χeE. Here χe is electric susceptibility.

1. Polarisation: P.
2. Electric Field: E.
3. Susceptibility: χe.
4. Formula Used: P = ε0χeE.
5. Final Result: Polarisation is proportional to electric field.

35. What is dielectric constant?

Dielectric constant is the factor by which capacitance increases when dielectric fills the space. It is K = C/C0.

1. Vacuum Capacitance: C0.
2. With Dielectric: C.
3. Formula Used: K = C/C0.
4. Final Result: Dielectric constant is dimensionless.

36. Why does a dielectric increase capacitance?

A dielectric increases capacitance by reducing electric field and potential difference. For the same charge, C = Q/V increases.

1. Polarisation: Produces opposing field.
2. Net Field: Decreases inside capacitor.
3. Potential Difference: Decreases for same Q.
4. Final Result: Dielectric increases capacitance by factor K.

Capacitors and Capacitance Class 12 Important Questions

A capacitor stores equal and opposite charges on two conductors.
The insulating gap prevents direct charge flow and maintains potential difference.
These capacitor class 12 physics questions cover definition, units, and practical meaning.

37. What is a capacitor?

A capacitor is a system of two conductors separated by an insulator. It stores charge and electrostatic energy.

1. Conductors: Two plates or conducting bodies.
2. Insulator: Air, vacuum, or dielectric.
3. Charge Pattern: +Q and −Q.
4. Final Result: A capacitor stores charge through potential difference.

38. What is capacitance?

Capacitance is the ratio of charge on one conductor to potential difference. Its formula is C = Q/V.

1. Charge: Q.
2. Potential Difference: V.
3. Formula Used: C = Q/V.
4. SI Unit: Farad.
5. Final Result: Capacitance measures charge storage ability.

39. What does one farad mean?

One farad means one coulomb charge produces one volt potential difference. It is a large practical unit.

1. Definition: C = Q/V.
2. If Q = 1 C and V = 1 V: C = 1 F.
3. Common Units: μF, nF, pF.
4. Final Result: 1 F = 1 C/V.

40. Why is farad a large unit?

Farad is large because ordinary capacitors store small charge at practical voltages. Common values use microfarad and picofarad.

1. Practical Capacitors: Often much less than 1 F.
2. Sub-units: 1 μF = 10^-6 F.
3. Small Unit: 1 pF = 10^-12 F.
4. Final Result: Farad is too large for many laboratory capacitors.

Parallel Plate Capacitor Class 12 Questions

A parallel plate capacitor uses two large conducting plates separated by a small distance.
Its field remains nearly uniform away from edges when plate area is large.
These parallel plate capacitor class 12 questions cover capacitance, dielectric insertion, and charge.

41. What is capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is C = ε0A/d in vacuum. It depends on area and separation.

1. Plate Area: A.
2. Separation: d.
3. Permittivity: ε0.
4. Formula Used: C = ε0A/d.
5. Final Result: Parallel plate capacitance equals ε0A/d.

42. How does area affect capacitance?

Capacitance increases when plate area increases. Larger area stores more charge for the same potential difference.

1. Formula Used: C = ε0A/d.
2. Area Relation: C ∝ A.
3. If A Doubles: C doubles.
4. Final Result: Capacitance is directly proportional to plate area.

43. How does separation affect capacitance?

Capacitance decreases when plate separation increases. The relation is inverse.

1. Formula Used: C = ε0A/d.
2. Separation Relation: C ∝ 1/d.
3. If d Doubles: C becomes half.
4. Final Result: Capacitance is inversely proportional to separation.

44. A capacitor has A = 6 × 10^-3 m² and d = 3 mm. Find capacitance.

The capacitance is 17.7 pF. Use C = ε0A/d.

1. Given Data:
   A = 6 × 10^-3 m²
   d = 3 mm = 3 × 10^-3 m
   ε0 = 8.85 × 10^-12 F/m
2. Formula Used: C = ε0A/d.
3. Calculation:
   C = 8.85 × 10^-12 × 6 × 10^-3/(3 × 10^-3)
   C = 17.7 × 10^-12 F
4. Final Result: C = 17.7 pF.

45. If this capacitor connects to 100 V, find charge on each plate.

The charge is 1.77 × 10^-9 C. Use Q = CV.

1. Given Data:
   C = 17.7 × 10^-12 F
   V = 100 V
2. Formula Used: Q = CV.
3. Calculation:
   Q = 17.7 × 10^-12 × 100
   Q = 1.77 × 10^-9 C
4. Final Result: Q = 1.77 nC.

Capacitors in Series and Parallel Class 12 Questions

Capacitors combine differently from resistors in many circuits.
Series capacitors carry equal charge, while parallel capacitors share the same voltage.
These capacitors in series and parallel class 12 questions cover equivalent capacitance and charge.

46. What is equivalent capacitance of capacitors in series?

For capacitors in series, 1/C = 1/C1 + 1/C2 + 1/C3 + .... Charge is same on each capacitor.

1. Series Condition: Same charge Q.
2. Voltage: Total voltage adds.
3. Formula Used: 1/C = 1/C1 + 1/C2 + ...
4. Final Result: Series capacitance is smaller than each capacitor.

47. What is equivalent capacitance of capacitors in parallel?

For capacitors in parallel, C = C1 + C2 + C3 + .... Voltage is same across each capacitor.

1. Parallel Condition: Same voltage V.
2. Charge: Total charge adds.
3. Formula Used: C = C1 + C2 + ...
4. Final Result: Parallel capacitance is the sum of capacitances.

48. Three 9 pF capacitors are connected in series. Find total capacitance.

The total capacitance is 3 pF. Equal series capacitors divide capacitance by number.

1. Given Data:
   C1 = C2 = C3 = 9 pF
2. Formula Used:
   1/C = 1/9 + 1/9 + 1/9
3. Calculation:
   1/C = 3/9 = 1/3
   C = 3 pF
4. Final Result: Total capacitance = 3 pF.

49. Three capacitors 2 pF, 3 pF, and 4 pF are in parallel. Find total capacitance.

The total capacitance is 9 pF. Parallel capacitances add directly.

1. Given Data:
   C1 = 2 pF
   C2 = 3 pF
   C3 = 4 pF
2. Formula Used: C = C1 + C2 + C3.
3. Calculation:
   C = 2 + 3 + 4
   C = 9 pF
4. Final Result: Total capacitance = 9 pF.

50. In parallel, find charge on 2 pF, 3 pF, and 4 pF capacitors at 100 V.

The charges are 200 pC, 300 pC, and 400 pC. Use Q = CV for each capacitor.

1. Given Data: V = 100 V.
2. Formula Used: Q = CV.
3. Calculations:
   Q1 = 2 pF × 100 V = 200 pC
   Q2 = 3 pF × 100 V = 300 pC
   Q3 = 4 pF × 100 V = 400 pC
4. Final Result: Charges are 200 pC, 300 pC, and 400 pC.

Energy Stored in Capacitor Class 12 Questions

A charged capacitor stores energy in the electric field between conductors.
This energy equals the work done in charging the capacitor.
These energy stored in capacitor class 12 questions cover formulas and numerical cases.

51. What is energy stored in a capacitor?

Energy stored in a capacitor is U = 1/2 CV². It also equals Q²/(2C) and 1/2 QV.

1. Formula 1: U = 1/2 CV².
2. Formula 2: U = Q²/(2C).
3. Formula 3: U = 1/2 QV.
4. Final Result: Stored energy has three equivalent forms.

52. What is electric field energy density?

Electric field energy density is u = 1/2 ε0E². It gives energy stored per unit volume.

1. Electric Field: E.
2. Permittivity: ε0.
3. Formula Used: u = 1/2 ε0E².
4. Final Result: Energy density depends on E².

53. A 12 pF capacitor connects to 50 V. Find stored energy.

The stored energy is 1.5 × 10^-8 J. Use U = 1/2 CV².

1. Given Data:
   C = 12 pF = 12 × 10^-12 F
   V = 50 V
2. Formula Used: U = 1/2 CV².
3. Calculation:
   U = 1/2 × 12 × 10^-12 × 50²
   U = 1.5 × 10^-8 J
4. Final Result: U = 1.5 × 10^-8 J.

54. A 900 pF capacitor is charged by 100 V. Find stored energy.

The stored energy is 4.5 × 10^-6 J. Use U = 1/2 CV².

1. Given Data:
   C = 900 pF = 900 × 10^-12 F
   V = 100 V
2. Formula Used: U = 1/2 CV².
3. Calculation:
   U = 1/2 × 900 × 10^-12 × 100²
   U = 4.5 × 10^-6 J
4. Final Result: U = 4.5 μJ.

55. Why is energy lost when a charged capacitor connects to an uncharged identical capacitor?

Energy is lost because transient current flows during charge sharing. The lost energy appears as heat and electromagnetic radiation.

1. Initial Capacitor: Charged.
2. Second Capacitor: Uncharged.
3. Connection: Charge redistributes.
4. Final Result: Final stored energy becomes half for identical capacitors.

NCERT Class 12 Physics Chapter 2 Questions

NCERT questions combine potential, conductors, capacitor combinations, dielectrics, and stored energy.
Students should convert cm, pF, μC, and voltage values before using formulas.
These NCERT Class 12 Physics Chapter 2 questions follow the 2026 textbook pattern.

56. Two charges 5 × 10^-8 C and −3 × 10^-8 C are 16 cm apart. Where is potential zero between them?

The potential is zero 10 cm from the positive charge between the charges. Use algebraic addition of potentials.

1. Given Data:
   q1 = 5 × 10^-8 C
   q2 = −3 × 10^-8 C
   Separation = 16 cm
2. Let Distance From Positive Charge: x.
3. Condition:
   5/x = 3/(16 − x)
4. Calculation:
   5(16 − x) = 3x
   80 − 5x = 3x
   x = 10 cm
5. Final Result: Zero potential point lies 10 cm from positive charge.

57. Find second zero-potential point for the same two charges.

The second zero-potential point lies 40 cm from the positive charge on the side of the negative charge.

1. Let Point Be Beyond Negative Charge: x cm from positive charge.
2. Distance From Negative Charge: x − 16.
3. Condition:
   5/x = 3/(x − 16)
4. Calculation:
   5x − 80 = 3x
   x = 40 cm
5. Final Result: The second point is 40 cm from the positive charge.

58. A regular hexagon of side 10 cm has 5 μC at each vertex. Find potential at centre.

The potential at the centre is 2.7 × 10^6 V. Each vertex is 10 cm from the centre.

1. Given Data:
   q = 5 μC = 5 × 10^-6 C
   r = 0.10 m
   Number of charges = 6
2. Formula Used: V = 6 × q/(4πε0r).
3. Calculation:
   V = 6 × 9 × 10^9 × 5 × 10^-6/0.10
   V = 2.7 × 10^6 V
4. Final Result: V = 2.7 × 10^6 V.

59. A spherical conductor of radius 12 cm has charge 1.6 × 10^-7 C. Find field inside it.

The electric field inside the spherical conductor is zero. Charges stay on its surface.

1. Conductor Condition: Electrostatic equilibrium.
2. Inside Region: No electric field.
3. Formula Result: E = 0.
4. Final Result: Field inside = 0 N/C.

60. Find electric field just outside the same spherical conductor.

The field just outside is 1.0 × 10^5 N/C. Use E = Q/(4πε0R²).

1. Given Data:
   Q = 1.6 × 10^-7 C
   R = 12 cm = 0.12 m
2. Formula Used: E = Q/(4πε0R²).
3. Calculation:
   E = 9 × 10^9 × 1.6 × 10^-7/(0.12)²
   E = 1.0 × 10^5 N/C
4. Final Result: E = 1.0 × 10^5 N/C outward.

61. An 8 pF capacitor has plate separation halved and dielectric K = 6 inserted. Find new capacitance.

The new capacitance is 96 pF. Halving distance doubles capacitance, and dielectric multiplies by 6.

1. Initial Capacitance: C0 = 8 pF.
2. Distance Halved: Capacitance becomes 2C0.
3. Dielectric Inserted: Capacitance becomes 6 × 2C0.
4. Calculation: C = 12 × 8 pF = 96 pF.
5. Final Result: New capacitance = 96 pF.

62. A 600 pF capacitor charged to 200 V connects to an uncharged 600 pF capacitor. Find energy lost.

The energy lost is 6 × 10^-6 J. For identical capacitors, final energy becomes half of initial energy.

1. Given Data:
   C = 600 pF = 600 × 10^-12 F
   V = 200 V
2. Initial Energy:
   Ui = 1/2 CV²
   Ui = 1/2 × 600 × 10^-12 × 200²
   Ui = 12 × 10^-6 J
3. Final Energy: Uf = Ui/2 = 6 × 10^-6 J.
4. Energy Lost: Ui − Uf = 6 × 10^-6 J.
5. Final Result: Energy lost = 6 μJ.

Class 12 Physics Chapter-Wise Important Questions