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Practice Set · Class 12

Class 12 Physics Numericals: Electrostatics & Current Electricity Practice Set

14 original problems, every numerical sub-type covered, every answer independently verified. Attempt each one before revealing the solution.

By Genelis Team 18 min read Updated 2026-08-22T09:00:00+05:30

Electrostatics and Current Electricity together carry some of the heaviest numerical weightage in Class 12 Physics — and both chapters cluster into a handful of recurring problem types once you've seen them mapped out. This is 14 original problems, one for every distinct sub-type across both chapters, each with a complete solution you can follow line by line.

How to use this page: Read each problem, attempt it fully on paper first, then tap "Reveal Solution" to check your working — not just your final answer. Every calculation on this page was computed independently and verified before publishing.

Every Sub-Type Covered — Across Both Chapters

E1

Coulomb's Law — force between charges

E2

Electric field due to a point charge

E3

Electric field on a dipole's axial line

E4

Potential energy of a charge system

E5

Field due to an infinite charged sheet

E6

Parallel plate capacitance & charge

E7

Capacitors in series & parallel + energy

C1

Resistance from resistivity & dimensions

C2

Series-parallel resistor combination

C3

Kirchhoff's laws — multi-loop circuit

C4

Wheatstone bridge balance condition

C5

EMF, internal resistance & terminal voltage

C6

Power dissipated in a resistor

C7

Potentiometer — comparing EMFs

Electrostatics

E1

Coulomb's Law

Two point charges of +4 μC and +9 μC are placed 30 cm apart in air. Calculate the electrostatic force between them. (k = 9 × 10⁹ N·m²/C²)

Reveal Solution
F = kq₁q₂/r²

Substitute

F = (9×10⁹ × 4×10⁻⁶ × 9×10⁻⁶) / (0.30)²

✓ F = 3.6 N (repulsive, since both charges are positive)
⚠️ Common mistake: forgetting to convert cm to metres before squaring — using 30 instead of 0.30 gives an answer off by a factor of 10,000.
E2

Electric Field — Point Charge

Calculate the electric field intensity at a point 20 cm from a point charge of +5 μC.

Reveal Solution
E = kq/r²

Substitute

E = (9×10⁹ × 5×10⁻⁶) / (0.20)²

✓ E = 1.125 × 10⁶ N/C, directed radially outward (away from the positive charge)
E3

Dipole — Axial Field

An electric dipole has a dipole moment of 4 × 10⁻⁸ C·m. Calculate the electric field at a point 10 cm from the centre of the dipole, on its axial line.

Reveal Solution

Identify the formula

For a short dipole, the axial field (at distance d, where d is large compared to the dipole's own length) is given by:

Eaxial = 2kp/d³

Substitute

E = (2 × 9×10⁹ × 4×10⁻⁸) / (0.10)³

✓ E = 7.2 × 10⁵ N/C, directed along the dipole axis (from −q to +q)
⚠️ Common mistake: using the equatorial field formula (E = kp/d³, without the factor of 2) instead of the axial formula. Always check which line — axial or equatorial — the question specifies.
E4

Potential Energy of a Charge System

Three identical point charges of +2 μC each are placed at the corners of an equilateral triangle of side 10 cm. Calculate the total electrostatic potential energy of the system.

Reveal Solution

Identify the pairs

With three charges, there are 3 distinct pairs, and by symmetry every pair is separated by the same distance (10 cm), since all sides of an equilateral triangle are equal.

U = 3 × (kq²/a)

Substitute

U = 3 × (9×10⁹ × (2×10⁻⁶)²) / 0.10

✓ U = 1.08 J
⚠️ Common mistake: calculating only one pair's energy and forgetting to multiply by 3 for all three pairs — or double-counting by using 6 instead of 3.
E5

Field Due to an Infinite Sheet (Gauss's Law)

An infinite plane sheet carries a uniform surface charge density of 2 μC/m². Calculate the electric field at a point near the sheet.

Reveal Solution

Apply the Gauss's law result for an infinite sheet

Using a Gaussian pillbox symmetric about the sheet gives a field independent of distance from the sheet.

E = σ / (2ε₀)

Substitute

E = (2×10⁻⁶) / (2 × 8.854×10⁻¹²)

✓ E ≈ 1.13 × 10⁵ N/C, directed perpendicular to and away from the sheet on both sides
⚠️ Common mistake: using E = σ/ε₀ (the conductor-surface formula) instead of E = σ/(2ε₀) for a thin, non-conducting charged sheet — these are two different physical situations with different formulas.
E6

Parallel Plate Capacitor

A parallel plate capacitor has plates of area 200 cm² separated by 1 mm in air. Calculate (a) its capacitance, and (b) the charge stored when connected to a 200 V supply.

Reveal Solution

Step 1 — Convert units

A = 200 cm² = 200×10⁻⁴ m² = 0.02 m². d = 1 mm = 0.001 m.

C = ε₀A/d

Substitute

C = (8.854×10⁻¹² × 0.02) / 0.001

(a) Capacitance

C ≈ 1.77 × 10⁻¹⁰ F = 177 pF

Q = CV

(b) Substitute

Q = 1.77×10⁻¹⁰ × 200

✓ C ≈ 177 pF, Q ≈ 3.54 × 10⁻⁸ C = 35.4 nC
E7

Capacitors in Series & Parallel + Energy

Three capacitors of 2 μF, 3 μF, and 6 μF are connected such that the 3 μF and 6 μF are in series with each other, and this combination is in parallel with the 2 μF capacitor. Calculate (a) the total capacitance, and (b) the energy stored if the combination is connected to a 10 V supply.

Reveal Solution

Step 1 — Series combination first

Cseries = (C₂×C₃)/(C₂+C₃) = (3×6)/(3+6) = 2 μF

Step 2 — Add in parallel with the 2 μF capacitor

Ctotal = 2 μF + 2 μF = 4 μF

Step 3 — Energy stored

U = ½CV² = ½ × 4×10⁻⁶ × 10²
✓ Ctotal = 4 μF, U = 2 × 10⁻⁴ J = 0.2 mJ
⚠️ Common mistake: adding all three capacitances directly as if they were all in parallel, ignoring that two of them are actually in series first. Always resolve the sub-combination before combining with the rest.

Current Electricity

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C1

Resistance from Resistivity

A copper wire of length 10 m and cross-sectional area 2 mm² has a resistivity of 1.7 × 10⁻⁸ Ω·m. Calculate its resistance.

Reveal Solution

Convert units

A = 2 mm² = 2×10⁻⁶ m²

R = ρL/A

Substitute

R = (1.7×10⁻⁸ × 10) / (2×10⁻⁶)

✓ R = 0.085 Ω
C2

Series-Parallel Combination

A resistor of 4 Ω is connected in series with a parallel combination of 6 Ω and 12 Ω resistors. Calculate the total equivalent resistance.

Reveal Solution

Step 1 — Resolve the parallel combination first

Rparallel = (6×12)/(6+12) = 4 Ω

Step 2 — Add the series resistor

Rtotal = 4 Ω + 4 Ω
✓ Rtotal = 8 Ω
C3

Kirchhoff's Laws — Multi-Loop Circuit

In a circuit, a 10 V cell drives current I₁ through a 2 Ω resistor, and a 4 V cell drives current I₂ through a 4 Ω resistor; both currents meet at a junction and flow together as I₃ through a shared 6 Ω resistor back to both cells. Using I₁ + I₂ = I₃, find all three currents.

Reveal Solution

Step 1 — Write the loop equations

Loop 1: 2I₁ + 6I₃ = 10. Loop 2: 4I₂ + 6I₃ = 4. Junction rule: I₁ + I₂ = I₃.

Step 2 — Substitute I₁ = I₃ − I₂ into Loop 1

2(I₃−I₂) + 6I₃ = 10 → 8I₃ − 2I₂ = 10

Step 3 — Express I₂ from Loop 2 and substitute

I₂ = 1 − 1.5I₃. Substituting: 8I₃ − 2(1−1.5I₃) = 10 → 11I₃ = 12

I₃ = 12/11 ≈ 1.09 A

Step 4 — Back-substitute

I₂ = 1 − 1.5(1.09) ≈ −0.64 A. I₁ = I₃ − I₂ ≈ 1.09 − (−0.64) ≈ 1.73 A

✓ I₁ ≈ 1.73 A, I₂ ≈ −0.64 A, I₃ ≈ 1.09 A
⚠️ I₂ comes out negative — this doesn't mean an error. It means the actual current in that branch flows opposite to the direction originally assumed. The magnitude, 0.64 A, is still correct; only the arrow direction flips. Always check both loop equations with your final values to confirm — here, 2(1.73)+6(1.09)=10 ✓ and 4(−0.64)+6(1.09)=4 ✓.
C4

Wheatstone Bridge Balance

In a balanced Wheatstone bridge, the resistances in three arms are P = 4 Ω, Q = 6 Ω, and R = 9 Ω. Calculate the unknown resistance S.

Reveal Solution

Apply the balance condition

At balance, no current flows through the galvanometer, giving:

P/Q = R/S → S = RQ/P

Substitute

S = (9 × 6) / 4

✓ S = 13.5 Ω
C5

EMF, Internal Resistance & Terminal Voltage

A cell of EMF 12 V and internal resistance 0.5 Ω is connected to an external resistor of 5.5 Ω. Calculate (a) the current in the circuit, and (b) the terminal voltage of the cell.

Reveal Solution
I = E / (r + R)

(a) Substitute

I = 12 / (0.5 + 5.5)

I = 2 A

V = E − Ir

(b) Substitute

V = 12 − (2 × 0.5)

✓ I = 2 A, Vterminal = 11 V
⚠️ Common mistake: forgetting to subtract the internal drop entirely and reporting the terminal voltage as equal to the EMF — terminal voltage always drops below EMF when current flows, due to the cell's own internal resistance.
C6

Power Dissipated

A current of 2 A flows through a resistor of 10 Ω. Calculate the power dissipated in the resistor.

Reveal Solution
P = I²R

Substitute

P = (2)² × 10

✓ P = 40 W
C7

Potentiometer — Comparing EMFs

In a potentiometer experiment, a standard cell of known EMF 1.5 V gives a balance length of 340 cm. When replaced with a cell of unknown EMF, the balance length becomes 240 cm. Calculate the unknown EMF.

Reveal Solution

Apply the potentiometer principle

Since potential drop along the wire is uniform per unit length, EMF is directly proportional to balance length.

E₂/E₁ = l₂/l₁

Substitute

E₂ = 1.5 × (240/340)

✓ E₂ ≈ 1.06 V

Which of These 14 Types Would You Actually Get Right Cold?

Reading through 14 solved problems and being able to solve 14 fresh ones without the type labelled for you are different skills. The real exam test is recognising which sub-type a new question belongs to on sight.

What a Genelis weak area map looks like after working through Electrostatics & Current Electricity problem sets

Coulomb's Law & basic fields
85%
Series-parallel resistor networks
71%
Capacitor combinations & energy
52%
Kirchhoff's laws — multi-loop circuits
34%

Next session: Kirchhoff's laws (34%) — not more Coulomb's Law practice. Genelis tracks accuracy by sub-type, not just by chapter, so it knows exactly which of these 14 patterns needs more reps.

Genelis is an AI-powered personalized learning platform built on Adaptive Personalized Intelligence. The Genelis learning system generates fresh, unlabelled Physics problems across all 14 sub-types, tracks your accuracy on each specifically, and logs every wrong answer to your wrong-question notebook for reattempt.

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Step 4 Wrong Qs auto-logged
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Step 5 Reattempt that type
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Practise unlimited fresh Physics problems on Genelis — free →
💡 For the full formula reference and chapter strategy behind these problem types, see the complete Class 12 Physics guide.
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Frequently Asked Questions

Questions Students Commonly Ask

Quick answers to the most common questions related to this guide.

What numerical types appear in CBSE Class 12 Electrostatics?

Seven distinct numerical sub-types recur across CBSE Class 12 Electrostatics: Coulomb's Law force calculation, electric field due to a point charge, electric field on the axial line of a dipole, potential energy of a system of point charges, electric field due to an infinite charged sheet using Gauss's law, parallel plate capacitance and charge calculation, and capacitors combined in series and parallel with energy stored.

What numerical types appear in CBSE Class 12 Current Electricity?

Seven recurring sub-types: resistance from resistivity and dimensions, equivalent resistance for series-parallel combinations, Kirchhoff's laws applied to multi-loop circuits, the Wheatstone bridge balance condition, EMF and internal resistance affecting terminal voltage, power dissipated in a resistor, and the potentiometer method for comparing EMFs.

What does it mean when a current comes out negative in a Kirchhoff's law problem?

A negative current value doesn't mean the calculation is wrong — it means the direction you initially assumed for that current was opposite to its actual direction in the circuit. The magnitude of the answer is still correct; you simply reverse the arrow you originally drew for that branch. This is a common, expected outcome in multi-loop Kirchhoff's law problems and is worth checking for specifically rather than assuming a negative sign is an error.

Why is the Wheatstone bridge balance condition written as P/Q = R/S?

At balance, no current flows through the galvanometer connecting the midpoints of the two arms, which means the potential at both midpoints is equal. This condition mathematically simplifies to the ratio of resistances in one arm equaling the ratio in the other arm: P/Q = R/S. This is what allows an unknown resistance to be calculated precisely once the bridge is balanced, without needing to know the exact current values anywhere in the circuit.

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