Electrostatics — Top 20 Questions

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Electric potential due to point charge \(5\times10^{-9}\) C is 50 V. Distance from charge:

A9 cm

B90 cm

C45 cm

D18 cm

✅ Correct: B

\(V = kq/r \Rightarrow r = kq/V = 9\times10^9\times5\times10^{-9}/50 = 45/50 = 0.9\text{ m} = \mathbf{90\text{ cm}}\).

Electric dipole moment \(6\times10^{-6}\) C·m in field \(1.5\times10^3\) N/C. Work in rotating 180°:

A9 mJ

B18 mJ

C36 mJ

✅ Correct: B

\(W = -p E(\cos\theta_f - \cos\theta_i) = -pE(\cos180° - \cos0°) = -pE(-1-1) = 2pE = 2\times6\times10^{-6}\times1500 = \mathbf{18\text{ mJ}}\).

Ratio of Coulomb force to gravitational force between electron and proton is of order:

A10¹⁹

B10³⁹

C10²⁹

D10³⁶

✅ Correct: B

\(F_e = ke^2/r^2\), \(F_g = Gm_em_p/r^2\). Ratio \(= ke^2/(Gm_em_p) = 9\times10^9\times(1.6\times10^{-19})^2/(6.67\times10^{-11}\times9.1\times10^{-31}\times1.67\times10^{-27}) \approx \mathbf{10^{39}}\).

Electric field on surface of uniformly charged spherical shell (surface charge density \(\sigma\)):

A\(\sigma/2\varepsilon_0\)

B\(\sigma/\varepsilon_0\)

C\(\sigma/4\varepsilon_0\)

D\(2\sigma/\varepsilon_0\)

✅ Correct: B

By Gauss's law: \(E \cdot 4\pi R^2 = Q/\varepsilon_0 = \sigma 4\pi R^2/\varepsilon_0\) → \(E = \mathbf{\sigma/\varepsilon_0}\). Inside shell: \(E = 0\). Just outside: \(E = \sigma/\varepsilon_0\).

Potential at centre of half ring (R=10 cm, linear charge density \(\lambda = 4\) nC/m) is \(x\pi\) V. Find \(x\):

A18

B36

C72

✅ Correct: B

\(V = k\lambda\pi R/R \cdot 1/\pi = k\lambda = ... \) Actually: \(V = k(\lambda\pi R)/R = k\lambda\pi = 9\times10^9\times4\times10^{-9}\times\pi = 36\pi\). So \(x = \mathbf{36}\).

Dimensions of \(\varepsilon_0(d\Phi_E/dt)\) same as:

AElectric charge

BElectric current

CElectric field

DElectric potential

✅ Correct: B

\(\varepsilon_0 d\Phi_E/dt\) = displacement current. It has dimensions of electric current (Amperes). Maxwell introduced this to complete Ampere's law.

Charges \(q\) and \(3q\) separated by \(r\). Distance from \(q\) where \(E = 0\):

A\(r/(1+\sqrt{3})\)

B\(r\sqrt{3}/(1+\sqrt{3})\)

C\(r/(1+1/\sqrt{3})\)

D\(r/2\)

✅ Correct: A

Let distance from \(q\) be \(x\): \(kq/x^2 = k3q/(r-x)^2\) → \((r-x)^2 = 3x^2\) → \(r-x = x\sqrt{3}\) → \(x = r/(1+\sqrt{3}) = \mathbf{r/(1+\sqrt{3})}\).

Energy stored in a capacitor \(C\) charged to voltage \(V\):

A\(CV\)

B\(CV^2\)

C\(CV^2/2\)

D\(2CV^2\)

✅ Correct: C

\(U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{QV}{2}\). The factor \(1/2\) comes from the fact that the average voltage during charging is \(V/2\). \(\mathbf{CV^2/2}\).

Three capacitors \(C_1, C_2, C_3\) in series across voltage \(V\). Charge on each is:

ASame

BDifferent

CZero for middle one

DDepends on capacitance

✅ Correct: A

In series combination, charge on each capacitor is the same. This is because the same charge \(Q\) is stored on each capacitor's plate as there is no external charge source between them.

Q10

Electric field inside a conductor in static equilibrium:

AEqual to surface charge density

BMaximum at centre

CZero

DProportional to voltage

✅ Correct: C

In electrostatic equilibrium, free charges redistribute to cancel any internal field. The electric field inside a conductor is always zero in static conditions.

Q11

Two equal and opposite charges \(\pm q\) are separated by \(2a\). Electric dipole moment:

A\(q/2a\)

B\(2qa\)

C\(qa\)

D\(4qa\)

✅ Correct: B

Electric dipole moment \(p = q\times d = q\times 2a = \mathbf{2qa}\). Direction: from negative to positive charge.

Q12

Capacitance of a parallel plate capacitor with area \(A\), separation \(d\), dielectric constant \(K\):

A\(\varepsilon_0 A/d\)

B\(K\varepsilon_0 A/d\)

C\(\varepsilon_0 A/Kd\)

D\(K^2\varepsilon_0 A/d\)

✅ Correct: B

\(C = \dfrac{K\varepsilon_0 A}{d}\). Without dielectric: \(C_0 = \varepsilon_0 A/d\). With dielectric (K>1), capacitance increases by factor \(K\). \(\mathbf{K\varepsilon_0 A/d}\).

Q13

Gauss's law: electric flux through a closed surface enclosing charge \(Q\):

A\(Q\varepsilon_0\)

B\(Q/4\pi\varepsilon_0\)

C\(Q/\varepsilon_0\)

D\(4\pi Q/\varepsilon_0\)

✅ Correct: C

Gauss's law: \(\Phi_E = \oint \vec{E}\cdot d\vec{A} = \dfrac{Q_\text{enc}}{\varepsilon_0} = \mathbf{Q/\varepsilon_0}\). This is independent of the shape and size of the Gaussian surface.

Q14

Work done in moving a charge \(q\) between two points at same electric potential:

A\(qV\)

B\(q/V\)

C\(V/q\)

✅ Correct: D

\(W = q(V_A - V_B)\). If \(V_A = V_B\), then \(W = 0\). Points at the same potential are called equipotential — no work is done moving a charge along an equipotential surface. \(\mathbf{W = 0}\).

Q15

Electric field due to infinite plane sheet of charge (surface charge density \(\sigma\)):

A\(\sigma/\varepsilon_0\)

B\(\sigma/2\varepsilon_0\)

C\(2\sigma/\varepsilon_0\)

D\(\sigma/4\varepsilon_0\)

✅ Correct: B

For infinite plane sheet: \(E = \dfrac{\sigma}{2\varepsilon_0}\) (on each side). For a conductor, \(E = \sigma/\varepsilon_0\) outside. Don't confuse — for isolated charged sheet: \(\mathbf{\sigma/2\varepsilon_0}\).

Q16

Potential energy of two charges \(q_1\) and \(q_2\) at separation \(r\):

A\(kq_1q_2/r^2\)

B\(kq_1q_2/r\)

C\(k(q_1+q_2)/r\)

D\(kq_1q_2r\)

✅ Correct: B

\(U = \dfrac{kq_1q_2}{r} = \dfrac{q_1q_2}{4\pi\varepsilon_0 r}\). This is positive for like charges (repulsion) and negative for unlike charges (attraction). \(\mathbf{kq_1q_2/r}\).

Q17

Van de Graaff generator works on principle of:

AElectromagnetic induction

BRedistribution of charge to outer surface of conductor

CCapacitor charging

DPiezoelectric effect

✅ Correct: B

Van de Graaff generator works because charge given to a conductor always resides on its outer surface. Charge is continuously transferred from belt to dome, building up very high potential.

Q18

Torque on a dipole (moment \(p\)) in uniform electric field \(E\) at angle \(\theta\):

A\(pE\sin\theta\)

B\(pE\cos\theta\)

C\(pE\tan\theta\)

D\(pE\)

✅ Correct: A

\(\vec{\tau} = \vec{p}\times\vec{E}\). Magnitude: \(\tau = pE\sin\theta\). Maximum at \(\theta = 90°\), zero at \(\theta = 0°\) and \(180°\). \(\mathbf{pE\sin\theta}\).

Q19

Force between two parallel plates of capacitor (opposite charges \(Q\), area \(A\)):

A\(Q^2/2A\varepsilon_0\)

B\(Q^2/A\varepsilon_0\)

C\(Q^2\varepsilon_0/A\)

D\(Q/A\varepsilon_0\)

✅ Correct: A

Electric field due to one plate \(= \sigma/2\varepsilon_0 = Q/2A\varepsilon_0\). Force on other plate \(= QE = Q\times Q/2A\varepsilon_0 = \mathbf{Q^2/2A\varepsilon_0}\).

Q20

Equivalent capacitance of three identical capacitors \(C\) in parallel:

A\(C/3\)

B\(3C\)

C\(C\)

D\(C^2/3\)

✅ Correct: B

In parallel: \(C_{eq} = C_1 + C_2 + C_3 = 3C\). Charge on each may differ but voltage is same. \(\mathbf{3C}\).

Roshan

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