Electric Charges and Fields Class 12 Physics Numericals

Electric Charges and Fields — Class 12 Physics Numericals
Step-by-step numerical practice on Coulomb's law, electric field, superposition, electric dipole, electric flux and Gauss's law.

How to Solve These Numericals

For every numerical, follow this order: Given → Formula → Substitution → Calculation → Unit → Final answer. Keep the sign and direction of a vector quantity clear wherever required.

1. Coulomb's Law

Numerical 1 — Force Between Two Charges

Question: Two point charges of +2 μC and −3 μC are placed 0.30 m apart in vacuum. Calculate the magnitude of the electrostatic force between them and state its nature.

Solution
q₁ = 2 × 10⁻⁶ C, q₂ = −3 × 10⁻⁶ C, r = 0.30 m
F = (1/4πε₀) |q₁q₂| / r²
F = 9 × 10⁹ × (2 × 10⁻⁶)(3 × 10⁻⁶) / (0.30)²
F = 0.60 N
Since the charges are unlike, the force is attractive.

Numerical 2 — Effect of Changing Distance

Question: The electrostatic force between two point charges is 36 N. What will the force become if their separation is increased to three times its original value?

Solution
By Coulomb's law, F ∝ 1/r².
If r' = 3r, then F' = F/9.
F' = 36/9 = 4 N.

Numerical 3 — Finding an Unknown Charge

Question: A charge of 4 μC experiences a force of 0.72 N due to another point charge placed 0.20 m away in vacuum. Find the magnitude of the other charge.

Solution
q₁ = 4 × 10⁻⁶ C, F = 0.72 N, r = 0.20 m
F = k|q₁q₂|/r²
q₂ = Fr²/(kq₁)
q₂ = 0.72 × 0.04 / (9 × 10⁹ × 4 × 10⁻⁶)
|q₂| = 0.8 μC.

2. Electric Field Due to a Point Charge

Numerical 4 — Field at a Point

Question: Calculate the electric field at a point 0.20 m from a point charge of +5 μC in vacuum.

Solution
E = kq/r²
E = 9 × 10⁹ × 5 × 10⁻⁶ / (0.20)²
E = 1.125 × 10⁶ N/C.
The field is radially outward because the source charge is positive.

Numerical 5 — Finding Distance from Electric Field

Question: At what distance from a point charge of 2 μC will the electric field have magnitude 4.5 × 10⁵ N/C?

Solution
E = kq/r²
r = √(kq/E)
r = √[(9 × 10⁹)(2 × 10⁻⁶)/(4.5 × 10⁵)]
r = 0.20 m.

3. Superposition of Electric Fields

Numerical 6 — Symmetric Charges

Question: Two identical charges +4 μC are placed symmetrically at equal distances from the origin on the x-axis. What is the resultant electric field at the origin?

Solution
Each charge produces an electric field at the origin of equal magnitude. Because the charges are identical and symmetrically placed, the two field vectors point in opposite directions.
Therefore, by superposition,
E_net = 0.

Numerical 7 — Unlike Symmetric Charges

Question: Equal charges +3 μC and −3 μC are placed symmetrically about the origin on the x-axis, each at distance 0.10 m from the origin. Find the magnitude and direction of the resultant electric field at the origin.

Solution
Field due to one charge:
E = kq/r² = (9 × 10⁹ × 3 × 10⁻⁶)/(0.10)²
E = 2.7 × 10⁶ N/C.
At the midpoint, the fields due to +q and −q point in the same direction, from the positive charge towards the negative charge.
E_net = 2E = 5.4 × 10⁶ N/C, directed from +3 μC toward −3 μC.

4. Electric Dipole

Numerical 8 — Dipole Moment

Question: An electric dipole consists of charges ±4 μC separated by 5 cm. Calculate its dipole moment.

Solution
p = q × d
= 4 × 10⁻⁶ × 5 × 10⁻²
p = 2 × 10⁻⁷ C m.

Numerical 9 — Torque on a Dipole

Question: An electric dipole of moment 4 × 10⁻⁸ C m is placed in a uniform electric field of 3 × 10⁵ N/C at an angle of 30°. Find the torque acting on it.

Solution
τ = pE sin θ
= (4 × 10⁻⁸)(3 × 10⁵) sin 30°
= 12 × 10⁻³ × 0.5
τ = 6 × 10⁻³ N m.

5. Electric Flux

Numerical 10 — Uniform Electric Field

Question: A uniform electric field of 3 × 10⁴ N/C passes normally through a plane surface of area 0.02 m². Calculate the electric flux.

Solution
Φ = EA cos θ
For normal incidence, θ = 0°, so cos θ = 1.
Φ = (3 × 10⁴)(0.02)
Φ = 600 N m²/C.

Numerical 11 — Inclined Surface

Question: A uniform electric field of 2 × 10⁴ N/C passes through a surface of area 0.50 m². The angle between the field and the area vector is 60°. Find the electric flux.

Solution
Φ = EA cos 60°
= (2 × 10⁴)(0.50)(0.5)
Φ = 5.0 × 10³ N m²/C.

6. Gauss's Law

Numerical 12 — Flux Through a Closed Surface

Question: A closed Gaussian surface encloses a charge of 6 μC. Calculate the total electric flux through the surface.

Solution
By Gauss's law,
Φ = q_enclosed/ε₀
= 6 × 10⁻⁶ / (8.85 × 10⁻¹²)
Φ ≈ 6.78 × 10⁵ N m²/C.

Numerical 13 — Infinite Line Charge

Question: An infinitely long straight wire has uniform linear charge density λ = 2 × 10⁻⁶ C/m. Calculate the electric field at a distance of 0.10 m from the wire.

Solution
For an infinite line charge,
E = λ/(2πε₀r)
= (2 × 10⁻⁶) / [2π(8.85 × 10⁻¹²)(0.10)]
E ≈ 3.60 × 10⁵ N/C.

Numerical 14 — Infinite Plane Sheet

Question: An infinite plane sheet has surface charge density σ = 4 × 10⁻⁸ C/m². Find the magnitude of the electric field near the sheet.

Solution
For a uniformly charged infinite plane sheet,
E = σ/(2ε₀)
= 4 × 10⁻⁸ / [2(8.85 × 10⁻¹²)]
E ≈ 2.26 × 10³ N/C.

Numerical 15 — Spherical Shell

Question: A thin spherical shell carries a total charge of 8 μC. What is the electric field at a point inside the shell? What is the field at an external point 0.40 m from the centre?

Solution
Inside the shell: A Gaussian surface entirely inside encloses no charge, so E = 0.

At 0.40 m outside:
E = kQ/r²
= (9 × 10⁹)(8 × 10⁻⁶)/(0.40)²
E = 4.5 × 10⁵ N/C.

7. Practice Without Solutions

Try these independently before checking your notes.

1. Two charges +5 μC and +2 μC are 0.50 m apart. Find the force between them.
2. A charge of 3 μC produces an electric field of 2.7 × 10⁵ N/C at a point. Find the distance of the point from the charge.
3. Two equal charges are separated by 20 cm. Find the electric field at the midpoint and explain the direction.
4. An electric dipole has charge magnitude 2 μC and separation 8 cm. Find its dipole moment.
5. A dipole of moment 5 × 10⁻⁷ C m is placed perpendicular to a field of 2 × 10⁴ N/C. Find the torque.
6. A surface of area 0.25 m² is placed at 60° to a uniform electric field of 4 × 10³ N/C. Find the flux if the angle is measured between E and the area vector.
7. A closed surface encloses 2 μC of charge. Find its total electric flux.
8. A long straight wire has λ = 5 × 10⁻⁷ C/m. Find the electric field at 20 cm.
9. An infinite plane sheet has σ = 1.77 × 10⁻⁸ C/m². Find the field near the sheet.
10. A uniformly charged spherical shell carries 5 μC. State the field at an interior point and calculate the field at an exterior point 0.50 m from the centre.

8. Formula Sheet for Numericals

Concept	Formula
Coulomb's law	F = (1/4πε0) |q1q2|/r²
Point charge field	E = (1/4πε0) q/r²
Dipole moment	p = qd
Dipole torque	τ = pE sin θ
Electric flux	Φ = EA cos θ
Gauss's law	Φ = qenclosed/ε0
Infinite line charge	E = λ/(2πε0r)
Infinite plane sheet	E = σ/(2ε0)
Useful constant	1/(4πε0) ≈ 9 × 109 N m²/C²

9. Exam Checklist

• Convert μC, nC and cm into SI units before substitution.
• Use vector direction carefully in superposition questions.
• For flux, check whether the angle is with the area vector or the surface.
• For Gauss's law, identify the enclosed charge before calculating flux.
• Choose a Gaussian surface that matches the symmetry of the charge distribution.
• Always write the final unit.

Curriculum basis: CBSE Class XII Physics 2026–27, Unit I: Electrostatics, Chapter 1: Electric Charges and Fields. The official syllabus includes Coulomb's law, superposition, electric field, electric dipole, electric flux and Gauss's theorem with applications to an infinite straight wire, infinite plane sheet and thin spherical shell. citeturn0search9