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Electrostatics — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Electrostatics MCQs with step-by-step solutions covering Coulomb's Law & Electric Field, Electric Dipole, Gauss' Law & Applications, Electric Potential & Potential Energy, Capacitors & Capacitance, Energy Stored & Dielectrics. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Coulomb's Law & Electric Field · easy · theory
Coulomb's law states that the force between two point charges is:
A. Inversely proportional to the product of the charges
B. Directly proportional to the separation between them
C. Directly proportional to the product of the charges and inversely proportional to the square of their separation ✓ Correct
D. Independent of the medium between them
Solution: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$, an inverse-square law like gravitation but vastly stronger.
Q2 — Coulomb's Law & Electric Field · easy · theory
The electrostatic force between two point charges acts:
A. At $45^\circ$ to the line joining them
B. Along the line joining the two charges ✓ Correct
C. Perpendicular to the line joining them
D. In a direction that depends on the medium
Solution: The Coulomb force is central: it is directed along the line of centres, repulsive for like charges and attractive for unlike ones.
Q3 — Coulomb's Law & Electric Field · easy · theory
The SI unit of electric field intensity is:
A. $\text{N}\cdot\text{C}$
B. $\text{N/C}$, equivalently $\text{V/m}$ ✓ Correct
C. $\text{J/C}$
D. $\text{C/N}$
Solution: Electric field is force per unit charge; the alternative form $\text{V/m}$ follows from $E = -\dfrac{dV}{dr}$.
Q4 — Coulomb's Law & Electric Field · easy · theory
The electric field inside the body of a charged conductor in electrostatic equilibrium is:
A. Directed radially inward
B. Zero ✓ Correct
C. Equal to the surface field
D. Maximum at the centre
Solution: Free charges redistribute themselves on the surface until the interior field vanishes; otherwise they would keep moving.
Q5 — Coulomb's Law & Electric Field · easy · numerical
If the distance between two point charges is doubled, the electrostatic force between them becomes:
A. One-half
B. One-fourth ✓ Correct
C. Twice
D. Four times
Solution: $F \propto \dfrac{1}{r^2}$, so doubling the separation reduces the force to a quarter.
Q6 — Coulomb's Law & Electric Field · easy · numerical
If both charges of a pair are doubled while their separation is unchanged, the force between them becomes:
A. Four times ✓ Correct
B. Twice
C. Unchanged
D. Half
Solution: $F \propto q_1q_2$, so doubling each charge multiplies the force by $4$.
Q7 — Coulomb's Law & Electric Field · easy · numerical
A charge of $2\,\mu\text{C}$ placed at a point experiences a force of $0.02\text{ N}$. The electric field there is:
A. $10^5\text{ N/C}$
B. $4 \times 10^{-8}\text{ N/C}$
C. $10^{-8}\text{ N/C}$
D. $10^4\text{ N/C}$ ✓ Correct
Solution: $E = \dfrac{F}{q} = \dfrac{0.02}{2 \times 10^{-6}} = 10^4\text{ N/C}$.
Q8 — Coulomb's Law & Electric Field · easy · numerical
The force between two charges in vacuum is $F$. When they are placed in a medium of dielectric constant $2$ at the same separation, the force becomes:
A. $4F$
B. $2F$
C. $\dfrac{F}{2}$ ✓ Correct
D. $\dfrac{F}{4}$
Solution: $F_{medium} = \dfrac{F}{K} = \dfrac{F}{2}$.
Q9 — Coulomb's Law & Electric Field · easy · numerical
A charge of $5\,\mu\text{C}$ is placed in a uniform electric field of $2 \times 10^4\text{ N/C}$. The force on it is:
A. $0.1\text{ N}$ ✓ Correct
B. $0.01\text{ N}$
C. $4 \times 10^9\text{ N}$
D. $1\text{ N}$
Solution: $F = qE = 5 \times 10^{-6} \times 2 \times 10^4 = 0.1\text{ N}$.
Q10 — Electric Dipole · easy · theory
The electric dipole moment of a dipole of charges $\pm q$ separated by $2l$ is:
A. $q \times 2l$, directed from $+q$ to $-q$
B. $2ql^2$
C. $\dfrac{q}{2l}$
D. $q \times 2l$, directed from $-q$ to $+q$ ✓ Correct
Solution: By convention the dipole moment vector points from the negative to the positive charge, and its unit is the coulomb metre.
Q11 — Electric Dipole · easy · theory
The torque acting on an electric dipole of moment $\vec{p}$ placed in a uniform electric field $\vec{E}$ is:
A. $\vec{p} \times \vec{E}$ ✓ Correct
B. $\vec{E} \times \vec{p}$
C. Zero always
D. $\vec{p} \cdot \vec{E}$
Solution: Its magnitude is $pE\sin\theta$, greatest when the dipole is perpendicular to the field and zero when aligned.
Q12 — Electric Dipole · easy · theory
The net force on an electric dipole placed in a uniform electric field is:
A. Equal to $pE$
B. Equal to $qE$
C. Equal to $2qE$
D. Zero ✓ Correct
Solution: The forces on the two charges are equal and opposite, so they form a couple: there is a torque but no net translational force.
Q13 — Electric Dipole · easy · theory
The SI unit of electric dipole moment is:
A. $\text{N}/\text{C}$
B. $\text{N}\cdot\text{m}$
C. $\text{C}/\text{m}$
D. $\text{C}\cdot\text{m}$ ✓ Correct
Solution: Dipole moment is charge times separation, so its unit is the coulomb metre.
Q14 — Electric Dipole · easy · numerical
An electric dipole placed in a uniform electric field experiences:
A. Neither force nor torque
B. A net force but no net torque
C. A net torque but no net force ✓ Correct
D. Both a net force and a net torque
Solution: The equal and opposite forces cancel in sum but have different lines of action, producing a pure couple.
Q15 — Gauss' Law & Applications · easy · theory
Gauss' law states that the total electric flux through a closed surface equals:
A. The charge enclosed divided by $\varepsilon_0$ ✓ Correct
B. Zero in all cases
C. The total charge in the universe divided by $\varepsilon_0$
D. The charge enclosed multiplied by $\varepsilon_0$
Solution: $\oint \vec{E}\cdot d\vec{A} = \dfrac{q_{enc}}{\varepsilon_0}$, where only charges inside the surface contribute.
Q16 — Gauss' Law & Applications · easy · theory
The electric field at any point inside a uniformly charged conducting spherical shell is:
A. Zero ✓ Correct
B. $\dfrac{kQ}{R^2}$
C. Infinite at the centre
D. $\dfrac{kQ}{r^2}$
Solution: A Gaussian surface drawn inside encloses no charge, so the field must vanish everywhere within.
Q17 — Gauss' Law & Applications · easy · theory
The net electric flux through a closed surface that encloses no net charge is:
A. Infinite
B. Always negative
C. Zero ✓ Correct
D. Always positive
Solution: Whatever flux enters the surface must leave it, so the net flux is zero even if the field inside is not.
Q18 — Gauss' Law & Applications · easy · theory
Electric flux is a:
A. Vector quantity
B. Scalar quantity ✓ Correct
C. Dimensionless quantity
D. Tensor quantity
Solution: It is defined by the scalar product $\vec{E}\cdot d\vec{A}$, and its unit is $\text{N}\cdot\text{m}^2/\text{C}$.
Q19 — Gauss' Law & Applications · easy · numerical
A spherical shell carries a charge $Q$. The electric field at a point inside the shell is:
A. $\dfrac{kQ}{r^2}$
B. Zero ✓ Correct
C. Infinite
D. $\dfrac{kQ}{R^2}$
Solution: A Gaussian sphere drawn inside encloses no charge, so by symmetry the field is zero everywhere within.
Q20 — Gauss' Law & Applications · easy · numerical
A uniform field of $100\text{ N/C}$ passes normally through an area of $2\text{ m}^2$. The flux is:
A. $100\text{ N}\cdot\text{m}^2/\text{C}$
B. Zero
C. $50\text{ N}\cdot\text{m}^2/\text{C}$
D. $200\text{ N}\cdot\text{m}^2/\text{C}$ ✓ Correct
Solution: $\Phi = EA\cos 0^\circ = 100 \times 2 = 200\text{ N}\cdot\text{m}^2/\text{C}$.
Q21 — Electric Potential & Potential Energy · easy · theory
The electric potential at a point is defined as:
A. The charge per unit area
B. The energy stored per unit volume
C. The force per unit charge at that point
D. The work done per unit positive charge in bringing it from infinity to that point ✓ Correct
Solution: Its SI unit is the volt, equal to one joule per coulomb.
Q22 — Electric Potential & Potential Energy · easy · theory
The electric potential at distance $r$ from a point charge $q$ is:
A. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$
B. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ ✓ Correct
C. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r}$
D. $\dfrac{1}{4\pi\varepsilon_0}qr$
Solution: Unlike the field, the potential falls off as $\dfrac{1}{r}$ and is a scalar, so potentials add algebraically.
Q23 — Electric Potential & Potential Energy · easy · theory
The work done in moving a charge between two points on the same equipotential surface is:
A. Zero ✓ Correct
B. Equal to $qV$
C. Maximum
D. Negative
Solution: $W = q(V_2 - V_1) = 0$ because the potential difference vanishes.
Q24 — Electric Potential & Potential Energy · easy · theory
Electric potential is a:
A. Vector quantity
B. Tensor quantity
C. Dimensionless quantity
D. Scalar quantity ✓ Correct
Solution: Potentials due to several charges add as ordinary numbers, which often makes potential easier to compute than field.
Q25 — Electric Potential & Potential Energy · easy · numerical
The work done in moving a charge of $2\,\mu\text{C}$ through a potential difference of $100\text{ V}$ is:
A. $200\text{ J}$
B. $2 \times 10^{-4}\text{ J}$ ✓ Correct
C. $5 \times 10^{-8}\text{ J}$
D. $2 \times 10^{-6}\text{ J}$
Solution: $W = q\Delta V = 2 \times 10^{-6} \times 100 = 2 \times 10^{-4}\text{ J}$.
Q26 — Capacitors & Capacitance · easy · theory
The capacitance of a conductor is defined as:
A. The work done per unit charge
B. The charge stored per unit area
C. The potential per unit charge
D. The charge required to raise its potential by one volt ✓ Correct
Solution: $C = \dfrac{Q}{V}$, measured in farad. One farad is an enormous capacitance, so practical units are $\mu\text{F}$ and $\text{pF}$.
Q27 — Capacitors & Capacitance · easy · theory
The capacitance of a parallel plate capacitor with plate area $A$ and separation $d$ in vacuum is:
A. $\dfrac{\varepsilon_0A}{d}$ ✓ Correct
B. $\dfrac{\varepsilon_0d}{A}$
C. $\varepsilon_0Ad$
D. $\dfrac{A}{\varepsilon_0d}$
Solution: Bringing the plates closer or enlarging them increases the charge the capacitor can hold at a given voltage.
Q28 — Capacitors & Capacitance · easy · theory
For capacitors connected in series, the equivalent capacitance satisfies:
A. $C = C_1 + C_2 + \ldots$
B. $C = \sqrt{C_1C_2}$
C. $C = C_1C_2$
D. $\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \ldots$ ✓ Correct
Solution: In series each capacitor carries the same charge while the potential differences add, so the equivalent capacitance is less than the smallest one.
Q29 — Capacitors & Capacitance · easy · theory
For capacitors connected in parallel, the equivalent capacitance is:
A. $\dfrac{C_1C_2}{C_1 + C_2}$
B. $C_1 + C_2 + \ldots$ ✓ Correct
C. $\dfrac{1}{C_1} + \dfrac{1}{C_2} + \ldots$
D. $\sqrt{C_1C_2}$
Solution: All capacitors share the same potential difference while the charges add, so the capacitances simply sum.
Q30 — Capacitors & Capacitance · easy · theory
Introducing a dielectric of constant $K$ between the plates of a capacitor changes its capacitance to:
A. $C$
B. $K^2C$
C. $KC$ ✓ Correct
D. $\dfrac{C}{K}$
Solution: Polarisation of the dielectric reduces the effective field, allowing more charge to be stored at the same voltage.