Electric Dipole — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Electric Dipole MCQs with step-by-step solutions (11 questions). Part of Electric Charges and Fields. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Electric Dipole · easy · theory
Polar molecules are the molecules
A. having zero dipole moment
B. acquire a dipole moment only in the presence of electric field due to displacement of charges
C. acquire a dipole moment only when magnetic field is absent
D. having a permanent electric dipole moment ✓ Correct
Solution: In polar molecules the centres of positive and negative charge do not coincide, giving a permanent electric dipole moment (e.g. water).
Q2 — Electric Dipole · medium · theory
The electric field at a point on the equatorial plane at a distance r from the centre of a dipole having dipole moment $\vec{P}$ is given by (r ≫ separation of the two charges)
A. $E = \frac{P}{4\pi\varepsilon_0 r^3}$
B. $E = \frac{2P}{4\pi\varepsilon_0 r^3}$
C. $E = -\frac{P}{4\pi\varepsilon_0 r^2}$
D. $E = -\frac{P}{4\pi\varepsilon_0 r^3}$ ✓ Correct
Solution: On the equatorial plane the field is $E = \frac{P}{4\pi\varepsilon_0 r^3}$ in magnitude, directed antiparallel to $\vec{P}$ — hence the negative sign: $E = -\frac{P}{4\pi\varepsilon_0 r^3}$
Q3 — Electric Dipole · medium · theory
An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle θ with the direction of the field. Assuming that the potential energy of the dipole is zero when θ = 90°, the torque and the potential energy of the dipole will respectively be
A. $pE\sin\theta, -pE\cos\theta$ ✓ Correct
B. $pE\sin\theta, -2pE\cos\theta$
C. $pE\sin\theta, 2pE\cos\theta$
D. $pE\cos\theta, -pE\sin\theta$
Solution: Torque $\tau = pE\sin\theta$. Potential energy $U = \int_{\pi/2}^{\theta}pE\sin\theta\,d\theta = -pE\cos\theta$
Q4 — Electric Dipole · medium · numerical
An electric dipole is placed at an angle of 30° with an electric field intensity $2 \times 10^5$ N/C. It experiences a torque equal to 4 N·m. The charge on the dipole, if the dipole length is 2 cm, is
A. 8 mC
B. 2 mC ✓ Correct
C. 5 mC
D. 7 µC
Solution: $\tau = pE\sin\theta \Rightarrow 4 = p \times 2 \times 10^5 \times \sin 30°$
$p = 4 \times 10^{-5}$ C·m; with $p = q(2l)$ and $2l = 2 \times 10^{-2}$ m:
$q = \frac{4 \times 10^{-5}}{2 \times 10^{-2}} = 2 \times 10^{-3}$ C = 2 mC
Q5 — Electric Dipole · hard · numerical
Three point charges +q, −2q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0), respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are
A. $\sqrt{2}qa$ along +y direction
B. $\sqrt{2}qa$ along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0) ✓ Correct
C. $qa$ along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)
D. $\sqrt{2}qa$ along +x direction
Solution: The −2q at the origin with +q on the x-axis and +q on the y-axis form two dipoles, each of moment qa, along x and y. Their resultant is $\sqrt{(qa)^2 + (qa)^2} = \sqrt{2}qa$, directed along the line from the origin towards (a, a, 0).
Q6 — Electric Dipole · easy · numerical
An electric dipole of moment p is lying along a uniform electric field E. The work done in rotating the dipole by 90° is
A. $\sqrt{2}pE$
B. $\frac{pE}{2}$
C. $2pE$
D. $pE$ ✓ Correct
Solution: $W = pE(\cos\theta_1 - \cos\theta_2) = pE(\cos 0° - \cos 90°) = pE(1 - 0) = pE$
Q7 — Electric Dipole · medium · theory
An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively
A. 2qE and minimum
B. qE and pE
C. zero and minimum ✓ Correct
D. qE and maximum
Solution: In a uniform field the two equal and opposite forces cancel, so net force = zero. With $p$ parallel to $E$ ($\theta = 0$), $U = -pE\cos 0° = -pE$, which is the minimum (most negative) value.
Q8 — Electric Dipole · medium · theory
A point Q lies on the perpendicular bisector of an electric dipole of dipole moment p. If the distance of Q from the dipole is r (much larger than the size of the dipole) then electric field at Q is proportional to
A. $p^{-1}$ and $r^2$
B. $p$ and $r^{-2}$
C. $p^2$ and $r^{-3}$
D. $p$ and $r^{-3}$ ✓ Correct
Solution: On the perpendicular bisector (equatorial line): $E = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}$, so $E \propto p$ and $E \propto r^{-3}$
Q9 — Electric Dipole · easy · theory
The formation of a dipole is due to two equal and dissimilar point charges placed at a
A. short distance ✓ Correct
B. long distance
C. above each other
D. None of these
Solution: An electric dipole consists of a pair of equal and opposite point charges separated by a very small (short) distance.
Q10 — Electric Dipole · easy · theory
Intensity of an electric field (E) due to a dipole, depends on distance r as
A. $E \propto \frac{1}{r}$
B. $E \propto \frac{1}{r^2}$
C. $E \propto \frac{1}{r^3}$ ✓ Correct
D. $E \propto \frac{1}{r^4}$
Solution: Both on the axial line ($E = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$) and the equatorial line ($E = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}$), the field varies as $E \propto \frac{1}{r^3}$
Q11 — Electric Dipole · medium · numerical
An electric dipole, consisting of two opposite charges of $2 \times 10^{-6}$ C each separated by a distance 3 cm is placed in an electric field of $2 \times 10^5$ N/C. Torque on the dipole is
A. $12 \times 10^{-1}$ N·m
B. $12 \times 10^{-2}$ N·m
C. $12 \times 10^{-3}$ N·m ✓ Correct
D. $12 \times 10^{-4}$ N·m
Solution: $\tau = pE = q(2l)E = (2 \times 10^{-6})(3 \times 10^{-2})(2 \times 10^5) = 12 \times 10^{-3}$ N·m (with $\theta = 90°$)