NEET Electric Charges and Fields PYQs & Practice Questions

10 free NEET Physics questions on Electric Charges and Fields, starting with previous year questions (2019–2025). Tap “Show answer” to check each one. No login needed.

Topics: Coulomb's Law · Properties of Electric Charges · Electric Field Lines · Gauss's Law · Electric Dipole · Electric Field due to Continuous Charge Distribution · Electric Dipole Field · Gauss's Law Applications · Basic Properties of Electric Charge · Charge Sharing by Conduction · Electric Field due to Point Charge · Electric Field due to Dipole

1. Q1NEET 2025Charge Sharing by Conduction

   Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (radii negligible compared to separation) is best given as :

   • A. 3F/5
   • B. 2F/3
   • C. F/2
   • D. 3F/8
   Show answer

   Answer: C — F/2

   Contact with A (charge q, uncharged C): both get q/2. Then C (q/2) contacts B (charge q): total 3q/2 splits equally, each gets 3q/4. Final charges: A = q/2, B = 3q/4. F' = k(q/2)(3q/4)/r² = (3/8)(kq²/r²) = (3/8)F.

2. Q2NEET 2025Electric Dipole in Uniform Field

   An electric dipole with dipole moment 5×10⁻⁶ Cm is aligned with the direction of a uniform electric field of magnitude 4×10⁵ N/C. The dipole is then rotated through an angle of 60° with respect to the electric field. The change in the potential energy of the dipole is :

   • A. 0.8 J
   • B. 1.0 J
   • C. 1.2 J
   • D. 1.5 J
   Show answer

   Answer: A — 0.8 J

   U1 = −pE cos0° = −5×10⁻⁶×4×10⁵ = −2 J. U2 = −pE cos60° = −2×(1/2) = −1 J. ΔU = U2−U1 = −1−(−2) = 1 J.

3. Q3NEET 2023Electric Dipole - Torque Calculation

   An electric dipole is placed at an angle of 30° with an electric field of intensity 2×10^5 NC^-1. It experiences a torque equal to 4 N m. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.

   • A. 6 mC
   • B. 4 mC
   • C. 2 mC
   • D. 8 mC
   Show answer

   Answer: B — 4 mC

   Torque τ = pE sinθ = q(2l)E sinθ. So 4 = q × 2×10^-2 × 2×10^5 × 0.5, giving 4/2 = q × 10^3, so q = 2×10^-3 C = 2 mC.

4. Q4NEET 2023Gauss's Law - Zero Net Flux

   If ∮E·ds = 0 over a surface, then:

   • A. the magnitude of electric field on the surface is constant.
   • B. all the charges must necessarily be inside the surface.
   • C. the electric field inside the surface is necessarily uniform.
   • D. the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
   Show answer

   Answer: C — the electric field inside the surface is necessarily uniform.

   If the net flux through a closed surface is zero, it means the number of field lines entering the surface equals the number leaving it, as per Gauss's law (net enclosed charge could also be zero, but this doesn't guarantee field uniformity or constant magnitude).

Practise all 25 Electric Charges and Fields questions with a timer

• Timed chapter test
• Score + weak topics
• Full NEET mocks & rank

Start free for 30 days

No payment, no card. Register in 2 minutes.

5. Q5NEET 2019Electric Field due to Charged Sphere

   A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre:

   • A. Zero as r increases from r < R increases as r increases for r > R
   • B. decreases as r increases for r < R and for r > R
   • C. increases as r increases for r < R and for r > R
   • D. zero as r increases for r < R, decreases as r increases for r > R
   Show answer

   Answer: C — increases as r increases for r < R and for r > R

   For a hollow charged sphere, inside the sphere (r<R) the electric field is zero everywhere (no charge enclosed). Outside the sphere (r>R), field E=kQ/r^2, which decreases as r increases.

6. Q6NEET 2019Electric Field due to Line Charges

   Two parallel infinite line charges with linear charge densities + lambda C/m and -lambda C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?

   • A. lambda/(pi*epsilon0*R) N/C
   • B. lambda/(2*pi*epsilon0*R) N/C
   • C. zero
   • D. 2*lambda/(pi*epsilon0*R) N/C
   Show answer

   Answer: —

   Each line charge produces field E = 2k*lambda/R at the midpoint (distance R from each line). Since fields from both line charges point in the same direction at the midpoint (one repels, one attracts), Enet = 2E = 2 x 2k*lambda/R = 4 x lambda/(4*pi*epsilon0*R) = lambda/(pi*epsilon0*R).

7. Q7Coulomb's Law

   Two point charges of +2 μC and +8 μC are placed 12 cm apart in vacuum. At what distance from the +2 μC charge, along the line joining them, is the electric field zero?

   • A. 4 cm
   • B. 6 cm
   • C. 8 cm
   • D. 3 cm
   Show answer

   Answer: —

   Let the field be zero at distance x from +2 μC charge. Then k(2)/x² = k(8)/(12-x)². Taking square root: √2/x = √8/(12-x) → (12-x)/x = 2 → 12-x = 2x → x = 4 cm. The point lies between the charges, 4 cm from the +2 μC charge.

8. Q8Properties of Electric Charges

   A glass rod rubbed with silk acquires a charge of +1.6 × 10⁻¹³ C. The number of electrons transferred from glass to silk is:

   • A. 1.6 × 10⁶
   • B. 1 × 10⁶
   • C. 1 × 10⁻³²
   • D. 2.56 × 10⁻³²
   Show answer

   Answer: A — 1.6 × 10⁶

   Number of electrons = Q/e = (1.6 × 10⁻¹³)/(1.6 × 10⁻¹⁹) = 10⁶ = 1 × 10⁶. These electrons are transferred from glass to silk, leaving glass positively charged.

Practise all 25 Electric Charges and Fields questions with a timer

• Timed chapter test
• Score + weak topics
• Full NEET mocks & rank

Start free for 30 days

No payment, no card. Register in 2 minutes.

9. Q9Electric Field Lines

   Which of the following statements about electric field lines is INCORRECT?

   • A. Electric field lines start from positive charges and end at negative charges.
   • B. Two electric field lines can intersect each other at a point where the field is very strong.
   • C. Electric field lines are closer together where the field is stronger.
   • D. Electric field lines do not form closed loops for electrostatic fields.
   Show answer

   Answer: A — Electric field lines start from positive charges and end at negative charges.

   Electric field lines can never intersect each other. If two lines intersected, it would mean there are two directions of electric field at that point, which is physically impossible. All other statements are correct properties of electric field lines.

10. Q10Gauss's Law

    A spherical shell of radius R has a surface charge density σ. The electric field just inside the shell is:

    • A. σ/ε₀
    • B. σ/2ε₀
    • C. Zero
    • D. σ/4ε₀
    Show answer

    Answer: B — σ/2ε₀

    By Gauss's law, the electric field inside a uniformly charged spherical shell is zero. The total charge enclosed by any Gaussian surface inside the shell is zero, so the net electric flux is zero, leading to E = 0. Just outside the shell, E = σ/ε₀.