Electric Dipole — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Electric Dipole MCQs with step-by-step solutions (21 questions). Part of Electrostatics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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.
Q2 — Electric Dipole · medium · theory
The electric field at an axial point of a short dipole varies with distance as:
A. $\dfrac{1}{r^4}$
B. $\dfrac{1}{r^2}$
C. $\dfrac{1}{r}$
D. $\dfrac{1}{r^3}$ ✓ Correct
Solution: $E_{axial} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$ — it falls off faster than a point charge because the two charges nearly cancel.
Q3 — Electric Dipole · medium · theory
The electric field at an equatorial point of a short dipole is:
A. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$
B. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$
C. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}$ ✓ Correct
D. Zero
Solution: The equatorial field is antiparallel to $\vec{p}$ and half the axial value at the same distance.
Q4 — Electric Dipole · medium · theory
At equal distances from a short dipole, the axial field compared with the equatorial field is:
A. Equal
B. Four times as large
C. Half as large
D. Twice as large ✓ Correct
Solution: Comparing $\dfrac{2p}{r^3}$ with $\dfrac{p}{r^3}$ shows the axial field is twice the equatorial one.
Q5 — 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.
Q6 — Electric Dipole · medium · theory
The potential energy of a dipole of moment $p$ in a uniform field $E$ at an angle $\theta$ is:
A. $-pE\cos\theta$ ✓ Correct
B. $pE\tan\theta$
C. $-pE\sin\theta$
D. $+pE\cos\theta$
Solution: $U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$, taking the zero of energy at $\theta = 90^\circ$.
Q7 — Electric Dipole · medium · theory
An electric dipole is in stable equilibrium in a uniform field when the angle between $\vec{p}$ and $\vec{E}$ is:
A. $45^\circ$
B. $0^\circ$ ✓ Correct
C. $180^\circ$
D. $90^\circ$
Solution: At $\theta = 0$ the potential energy $-pE$ is a minimum, so any displacement produces a restoring torque.
Q8 — 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.
Q9 — 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.
Q10 — Electric Dipole · medium · numerical
A dipole consists of charges $\pm 2\,\mu\text{C}$ separated by $2\text{ cm}$. Its dipole moment is:
A. $2 \times 10^{-8}\text{ C}\cdot\text{m}$
B. $10^{-4}\text{ C}\cdot\text{m}$
C. $4 \times 10^{-8}\text{ C}\cdot\text{m}$ ✓ Correct
D. $4 \times 10^{-6}\text{ C}\cdot\text{m}$
Solution: $p = q \times 2l = 2 \times 10^{-6} \times 0.02 = 4 \times 10^{-8}\text{ C}\cdot\text{m}$.
Q11 — Electric Dipole · medium · numerical
A dipole of moment $4 \times 10^{-8}\text{ C}\cdot\text{m}$ is placed in a uniform field of $10^5\text{ N/C}$. The maximum torque on it is:
A. $4 \times 10^{3}\text{ N}\cdot\text{m}$
B. $2 \times 10^{-3}\text{ N}\cdot\text{m}$
C. $4 \times 10^{-3}\text{ N}\cdot\text{m}$ ✓ Correct
D. $4 \times 10^{-13}\text{ N}\cdot\text{m}$
Solution: Maximum torque occurs at $\theta = 90^\circ$: $\tau = pE = 4 \times 10^{-8} \times 10^5 = 4 \times 10^{-3}\text{ N}\cdot\text{m}$.
Q12 — Electric Dipole · medium · numerical
A dipole of moment $4 \times 10^{-8}\text{ C}\cdot\text{m}$ makes $30^\circ$ with a field of $10^5\text{ N/C}$. The torque on it is:
A. Zero
B. $3.46 \times 10^{-3}\text{ N}\cdot\text{m}$
C. $2 \times 10^{-3}\text{ N}\cdot\text{m}$ ✓ Correct
D. $4 \times 10^{-3}\text{ N}\cdot\text{m}$
Solution: $\tau = pE\sin 30^\circ = 4 \times 10^{-3} \times 0.5 = 2 \times 10^{-3}\text{ N}\cdot\text{m}$.
Q13 — Electric Dipole · medium · numerical
The potential energy of a dipole placed perpendicular to a uniform electric field is:
A. $+pE$
B. $\dfrac{pE}{2}$
C. $-pE$
D. Zero ✓ Correct
Solution: $U = -pE\cos 90^\circ = 0$, which is the conventional reference position.
Q14 — Electric Dipole · medium · numerical
The potential energy of a dipole aligned antiparallel to a uniform field $E$ is:
A. $+pE$ ✓ Correct
B. Zero
C. $\dfrac{pE}{2}$
D. $-pE$
Solution: $U = -pE\cos 180^\circ = +pE$ — the position of maximum energy and unstable equilibrium.
Q15 — Electric Dipole · hard · numerical
The work done in rotating a dipole from $\theta = 0^\circ$ to $\theta = 90^\circ$ in a uniform field $E$ is:
A. $\dfrac{pE}{2}$
B. Zero
C. $2pE$
D. $pE$ ✓ Correct
Solution: $W = U_{90^\circ} - U_{0^\circ} = 0 - (-pE) = pE$.
Q16 — Electric Dipole · hard · numerical
The work done in rotating a dipole from $\theta = 0^\circ$ to $\theta = 180^\circ$ in a uniform field $E$ is:
A. $\dfrac{pE}{2}$
B. Zero
C. $2pE$ ✓ Correct
D. $pE$
Solution: $W = (+pE) - (-pE) = 2pE$ — the maximum possible work.
Q17 — Electric Dipole · hard · numerical
A short dipole of moment $10^{-8}\text{ C}\cdot\text{m}$ produces an axial field at $0.1\text{ m}$ of:
A. $1.8 \times 10^5\text{ N/C}$
B. $1.8 \times 10^4\text{ N/C}$ ✓ Correct
C. $9 \times 10^3\text{ N/C}$
D. $9 \times 10^5\text{ N/C}$
Solution: $E = \dfrac{2kp}{r^3} = \dfrac{2 \times 9 \times 10^9 \times 10^{-8}}{10^{-3}} = \dfrac{180}{10^{-3}} = 1.8 \times 10^5\text{ N/C}$.
Q18 — Electric Dipole · hard · numerical
For the dipole of the previous type ($p = 10^{-8}\text{ C}\cdot\text{m}$), the equatorial field at $0.1\text{ m}$ is:
A. $1.8 \times 10^5\text{ N/C}$
B. $9 \times 10^4\text{ N/C}$ ✓ Correct
C. $9 \times 10^3\text{ N/C}$
D. $4.5 \times 10^4\text{ N/C}$
Solution: $E_{eq} = \dfrac{kp}{r^3} = \dfrac{9 \times 10^9 \times 10^{-8}}{10^{-3}} = 9 \times 10^4\text{ N/C}$, exactly half the axial value.
Q19 — Electric Dipole · medium · numerical
If the distance from a short dipole is doubled, the axial electric field becomes:
A. One-eighth ✓ Correct
B. One-half
C. One-fourth
D. One-sixteenth
Solution: $E \propto \dfrac{1}{r^3}$, so doubling $r$ divides the field by $2^3 = 8$.
Q20 — 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.
Q21 — Electric Dipole · medium · numerical
A dipole of moment $2 \times 10^{-8}\text{ C}\cdot\text{m}$ is held perpendicular to a uniform field of $5 \times 10^4\text{ N/C}$. The torque on it is:
A. Zero
B. $5 \times 10^{-4}\text{ N}\cdot\text{m}$
C. $2 \times 10^{-3}\text{ N}\cdot\text{m}$
D. $10^{-3}\text{ N}\cdot\text{m}$ ✓ Correct
Solution: $\tau = pE\sin 90^\circ = 2 \times 10^{-8} \times 5 \times 10^4 = 10^{-3}\text{ N}\cdot\text{m}$.