Pravegaa PYQ Series · Volume 3: Electromagnetic Theory, Waves & Optics

EM02: Boundary-Value Problems and Multipoles

49 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

IIT JAM: 4
GATE: 8
CSIR NET: 7
JEST: 13
TIFR GS: 17

EM02.1IIT JAM 2025, Q6MCQ | 1 mark

Consider a parallel plate capacitor (distance between the plates \(d\), and permittivity \(\varepsilon_0\)) as shown in the figure below. The space charge density between the plates varies as \(\rho(x)=\rho_0e^{-x}\). Voltage \(V=0\) both at \(x=0\) and \(x=d\).The voltage \(V(x)\) at point P between the plates is

[\(\rho_0\) is a constant of appropriate dimensions]

\(\dfrac{\rho_0}{\varepsilon_0}\left[e^{-x}+\dfrac{1-e^{-d}}{d}x-1\right]\)
\(\dfrac{2\rho_0}{\varepsilon_0}\left[e^{-x}+\dfrac{1-e^{-d}}{d}x-1\right]\)
\(\dfrac{\rho_0}{2\varepsilon_0}\left[e^{-x}+\dfrac{1-e^{-d}}{d}x-1\right]\)
\(\dfrac{3\rho_0}{\varepsilon_0}\left[e^{-x}+\dfrac{1-e^{-d}}{d}x-1\right]\)

Show solution

In one dimension Poisson's equation reads \[\frac{d^2V}{dx^2}=-\frac{\rho_0e^{-x}}{\varepsilon_0}\;\Rightarrow\;\frac{dV}{dx}=\frac{\rho_0e^{-x}}{\varepsilon_0}+c_1\;\Rightarrow\;V(x)=-\frac{\rho_0e^{-x}}{\varepsilon_0}+c_1x+c_2 .\] \(V(0)=0\) gives \(c_2=\rho_0/\varepsilon_0\). \(V(d)=0\) gives \[c_1d=\frac{\rho_0}{\varepsilon_0}\left(e^{-d}-1\right)\;\Rightarrow\;c_1=-\frac{\rho_0}{\varepsilon_0}\,\frac{1-e^{-d}}{d}.\] Therefore \[\boxed{V(x)=-\frac{\rho_0}{\varepsilon_0}\left[e^{-x}+\frac{1-e^{-d}}{d}\,x-1\right]}\] The \(x\)-dependence is the bracket \(\left[e^{-x}+\frac{1-e^{-d}}{d}x-1\right]\) common to all four options, and the prefactor has magnitude \(\rho_0/\varepsilon_0\). The other options carry prefactors \(2\rho_0/\varepsilon_0\), \(\rho_0/2\varepsilon_0\) and \(3\rho_0/\varepsilon_0\), so the answer is option (A).

Answer: (A)

EM02.2IIT JAM 2021, Q18MCQ | 2 marks

Four charges are placed very close to each other, as shown. The separation between the two charges on the \(y\)-axis is \(a\). The separation between the two charges on the \(x\)-axis is also \(a\). The leading order (non-vanishing) form of the electrostatic potential, at point \(P\), at a distance \(r\) from the origin \((r\gg a)\), is

\(\dfrac{1}{4\pi\epsilon_0}\dfrac{qa}{2r^2}\left(\sqrt3-1\right)\)
\(\dfrac{1}{4\pi\epsilon_0}\dfrac{2qa}{r^2}\)
\(\dfrac{1}{4\pi\epsilon_0}\dfrac{qa}{r^2}\left(\sqrt5-1\right)\)
\(\dfrac{1}{4\pi\epsilon_0}\dfrac{qa}{r^2}\left(1-\sqrt3\right)\)

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The net charge is zero, so the leading term is the dipole term. The charges are \(+q\) at \((0,\tfrac a2)\), \(-q\) at \((0,-\tfrac a2)\), \(+q\) at \((-\tfrac a2,0)\) and \(-q\) at \((\tfrac a2,0)\): \[ \vec p=\sum_i q_i\vec r_i=qa\,\hat y+(-qa\,\hat x)=qa\left(-\hat x+\hat y\right). \] For \(P\) at \(60^\circ\) to the \(x\)-axis, \(\hat r=\cos60^\circ\,\hat x+\sin60^\circ\,\hat y=\tfrac12\hat x+\tfrac{\sqrt3}{2}\hat y\), so \[ V=\frac{1}{4\pi\epsilon_0}\frac{\vec p\cdot\hat r}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{qa}{r^2}\left(-\frac12+\frac{\sqrt3}{2}\right) \] \[ \boxed{V=\frac{1}{4\pi\epsilon_0}\frac{qa}{2r^2}\left(\sqrt3-1\right)} \]

Answer: (A)

EM02.3IIT JAM 2017, Q34MSQ | 2 marks

For a point dipole of dipole moment \(\vec p=p\hat z\) located at the origin, which of the following is (are) correct?

(A) The electric field at \((0,b,0)\) is zero

(B) The work done in moving a charge \(q\) from \((0,b,0)\) to \((0,0,b)\) is \(\dfrac{qp}{4\pi\epsilon_0b^2}\)

(D) If a charge \(q\) is kept at \((0,0,b)\) it will exert a force of magnitude \(\dfrac{qp}{4\pi\epsilon_0b^3}\) on the dipole

Show solution

For the dipole, \[ V=\frac{p\cos\theta}{4\pi\epsilon_0r^2},\qquad \vec E=\frac{p}{4\pi\epsilon_0r^3}\left(2\cos\theta\,\hat r+\sin\theta\,\hat\theta\right). \] (A) At \((0,b,0)\), \(\theta=\pi/2\): \(\vec E=\dfrac{p}{4\pi\epsilon_0b^3}\hat\theta=-\dfrac{p}{4\pi\epsilon_0b^3}\hat z\neq0\). Incorrect.

(B) \(V(0,b,0)=0\) and \(V(0,0,b)=\dfrac{p}{4\pi\epsilon_0b^2}\), so \[ W=q\left[V(0,0,b)-V(0,b,0)\right]=\frac{qp}{4\pi\epsilon_0b^2}. \] Correct.

(D) The dipole's field at the axial point \((0,0,b)\) is \(\dfrac{2p}{4\pi\epsilon_0b^3}\hat z\), so the force on \(q\) is \(\dfrac{2qp}{4\pi\epsilon_0b^3}\). By Newton's third law the force on the dipole has the same magnitude, \(\dfrac{2qp}{4\pi\epsilon_0b^3}\). Incorrect. \[ \boxed{(B),\ (C)} \]

Answer: (B), (C)

EM02.4IIT JAM 2016, Q32MSQ | 2 marks

For an electric dipole with moment \(\vec P=p_0\hat e_z\) placed at the origin, (\(p_0\) is a constant of appropriate dimensions and \(\hat e_x,\hat e_y\) and \(\hat e_z\) are unit vectors in Cartesian coordinate system)

potential falls as \(\dfrac{1}{r^2}\), where \(r\) is the distance from origin
a spherical surface centered at origin is an equipotential surface
electric flux through a spherical surface enclosing the origin is zero
radial component of \(\vec E\) is zero on the \(xy\)-plane.

Show solution

The dipole potential and field are \[ V=\frac{p_0\cos\theta}{4\pi\varepsilon_0r^2},\qquad E_r=\frac{2p_0\cos\theta}{4\pi\varepsilon_0r^3},\qquad E_\theta=\frac{p_0\sin\theta}{4\pi\varepsilon_0r^3}. \]

\(V\propto1/r^2\): (A) is correct.
On a sphere \(r=R\), \(V\) still varies as \(\cos\theta\), so the sphere is not an equipotential: (B) is false.
The net charge of a dipole is zero, so by Gauss's law the flux through any closed surface enclosing it is \(Q_{\text{enc}}/\varepsilon_0=0\): (C) is correct.
On the \(xy\)-plane \(\theta=\pi/2\), so \(E_r\propto\cos\theta=0\): (D) is correct.

\[ \boxed{(A),\ (C),\ (D)} \]

Answer: (A), (C), (D)

EM02.5GATE 2026, Q48MCQ | 2 marks

A positive point charge is fixed at the origin. At some distance from it on the \(x\) axis, a point dipole is kept pointing in the \(+y\) direction. The force on the dipole is

0
in the \(+y\) direction
in the \(-y\) direction
in the \(+x\) direction

Show solution

\(\vec F=(\vec p\cdot\vec\nabla)\vec E=p\,\partial_y\vec E\) evaluated at \((d,0,0)\), with \(\vec E=kq\,\dfrac{(x,y,z)}{r^3}\): \[ \partial_yE_x=-\frac{3kqxy}{r^5}\bigg|_{y=0}=0,\qquad \partial_yE_y=kq\left(\frac1{r^3}-\frac{3y^2}{r^5}\right)\bigg|_{y=0}=\frac{kq}{d^3}. \] So \(\vec F=\dfrac{kqp}{d^3}\,\hat y\), along \(\boxed{+y}\). (Consistent with Newton's third law: the dipole's field at the origin is \(-\vec p/(4\pi\epsilon_0d^3)\), pushing the charge along \(-y\).)

Answer: (B) in the \(+y\) direction

EM02.6GATE 2025, Q40MCQ | 2 marks

A thin circular ring of radius \(R\) lies in the \(xy\) plane with its centre coinciding with the origin. The ring carries a uniform line charge density \(\lambda\). The quadrupole contribution to the electrostatic potential at the point \((0,0,d)\), where \(d\gg R\), is

\(-\dfrac{\lambda R^3}{4\epsilon_0d^3}\)
0
\(\dfrac{\lambda R^3}{4\epsilon_0d^3}\)
\(-\dfrac{\lambda R^3}{2\epsilon_0d^3}\)

Show solution

Total charge: \(Q=2\pi R\lambda\). The exact potential on the axis is \[V(d)=\frac{1}{4\pi\epsilon_0}\frac{Q}{\sqrt{d^2+R^2}} .\] For \(d\gg R\), using the binomial expansion, \[\frac{1}{\sqrt{d^2+R^2}}=\frac1d\left(1+\frac{R^2}{d^2}\right)^{-1/2}\approx\frac1d-\frac{R^2}{2d^3}+\cdots\] so \[V(d)\approx\frac{Q}{4\pi\epsilon_0d}-\frac{QR^2}{8\pi\epsilon_0d^3}+\cdots\] The first term is the monopole term; the dipole term vanishes by symmetry; the \(1/d^3\) term is the quadrupole contribution: \[V_{\text{quad}}=-\frac{(2\pi R\lambda)R^2}{8\pi\epsilon_0d^3}=-\frac{\lambda R^3}{4\epsilon_0d^3}.\] \[\boxed{V_{\text{quad}}=-\dfrac{\lambda R^3}{4\epsilon_0d^3}}\]

Answer: (A) \(-\dfrac{\lambda R^3}{4\epsilon_0 d^3}\)

EM02.7GATE 2025, Q49MSQ | 2 marks

A point charge \(q\) is placed at a distance \(d\) above an infinite, grounded conducting plate placed on the \(xy\) plane at \(z=0\). The electrostatic potential in \(z>0\) region is given by \(\phi=\phi_1+\phi_2\), where \[\phi_1=\frac{1}{4\pi\epsilon_0}\frac{q}{\sqrt{x^2+y^2+(z-d)^2}}\quad\text{and}\quad\phi_2=-\frac{1}{4\pi\epsilon_0}\frac{q}{\sqrt{x^2+y^2+(z+d)^2}}.\] Which of the following option(s) is/are correct?

The magnitude of the force experienced by the point charge \(q\) is \(\dfrac{1}{16\pi\epsilon_0}\dfrac{q}{d^2}\)
The electrostatic energy of the system is \(\dfrac{1}{8\pi\epsilon_0}\dfrac{q^2}{d}\)
The induced surface charge density on the plate is proportional to \(\dfrac{1}{\sqrt{x^2+y^2+d^2}}\)
The electrostatic potential \(\phi_1\) satisfies Poisson's equation for \(z>0\)

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(A) False. The force is that of the image charge \(-q\) at distance \(2d\): \[|F|=\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}=\frac{1}{16\pi\epsilon_0}\frac{q^2}{d^2},\] which involves \(q^2\), not \(q\).
(B) False. The energy is half the energy of the charge–image pair: \[W=-\frac{1}{16\pi\epsilon_0}\frac{q^2}{d}\] (negative, and with \(16\pi\epsilon_0\), not \(8\pi\epsilon_0\)).
(C) False. \(\sigma=-\epsilon_0\left.\dfrac{\partial\phi}{\partial z}\right|_{z=0}=-\dfrac{qd}{2\pi\left(x^2+y^2+d^2\right)^{3/2}}\), which varies as \((x^2+y^2+d^2)^{-3/2}\).
(D) True. \(\phi_1\) is the potential of the real charge at \((0,0,d)\), which lies in \(z>0\): \(\nabla^2\phi_1=-\dfrac{q}{\epsilon_0}\delta^3(\vec r-d\hat z)\), i.e. Poisson's equation.

\[\boxed{\text{(D)}}\]

Answer: (D)

EM02.8GATE 2025, Q56MSQ | 2 marks

A neutral conducting sphere of radius \(R\) is placed in a uniform electric field of magnitude \(E_0\), that points along the \(z\) axis. The electrostatic potential at any point \(\vec r\) outside the sphere is given by \(V(r,\theta)=V_0-E_0r\left(1-\dfrac{R^3}{r^3}\right)\cos\theta\), where \(V_0\) is the constant potential of the sphere. Which of the following option(s) is/are correct?

The induced surface charge density on the sphere is proportional to \(\sin\theta\)
As \(r\to\infty\), \(\vec E=E_0\cos\theta\,\hat r\)
The electric field at any point is curl free for \(r>R\)
The electric field at any point is divergence free for \(r>R\)

Show solution

For \(r>R\), \(V=V_0-E_0\left(r-\dfrac{R^3}{r^2}\right)\cos\theta\) and \(\vec E=-\nabla V\).

(A) False. \(\sigma=\epsilon_0E_r(R,\theta)=-\epsilon_0\left.\dfrac{\partial V}{\partial r}\right|_R=\epsilon_0E_0\left(1+\dfrac{2R^3}{R^3}\right)\cos\theta=3\epsilon_0E_0\cos\theta\propto\cos\theta\).
(B) False. As \(r\to\infty\), \(\vec E\to E_0\hat z=E_0\cos\theta\,\hat r-E_0\sin\theta\,\hat\theta\), which is not purely radial.
(C) True. An electrostatic field is a gradient, so \(\nabla\times\vec E=0\).
(D) True. There is no charge in \(r>R\), so \(\nabla\cdot\vec E=\rho/\epsilon_0=0\) (equivalently \(\nabla^2V=0\) there).

\[\boxed{\text{(C), (D)}}\]

Answer: (C), (D)

EM02.9GATE 2024, Q46MSQ | 2 marks

Two-point charges of charge \(+q\) each are placed a distance \(2d\) apart. A grounded solid conducting sphere of radius \(a\) is placed midway between them. Assume \(a^2\ll d^2\). Which of the following statement is/are true?

If \(a>\frac d8\), the net force acting on the charges is directed towards each other
The potential at the surface of the sphere is zero
Total induced charge on the sphere is \(\left(-\frac{2aq}{d}\right)\)
The potential at the center of the sphere is non-zero

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For a grounded sphere, each charge \(q\) at distance \(d\) from the centre has an image \(q'=-\dfrac{a}{d}q\) at distance \(\dfrac{a^2}{d}\) from the centre on the same side.

(B) The sphere is grounded, so its surface (and the whole conductor) is at zero potential. True; for the same reason the potential at the centre is zero, so (D) is false.

(A) Consider the right-hand charge. The other charge repels it with \(\dfrac{1}{4\pi\epsilon_0}\dfrac{q^2}{(2d)^2}\) (outward). The two images, at distances \(d-\frac{a^2}{d}\approx d\) and \(d+\frac{a^2}{d}\approx d\) (since \(a^2\ll d^2\)), attract it with \[\frac{1}{4\pi\epsilon_0}\,\frac{2(aq/d)\,q}{d^2}=\frac{1}{4\pi\epsilon_0}\frac{2aq^2}{d^3}\ \text{(inward)}.\] The net force is inward (charges pulled towards each other) if \[\frac{2a}{d^3}>\frac{1}{4d^2}\;\Rightarrow\;a>\frac d8 .\] True. \[\boxed{\text{(A), (B), (C)}}\]

Answer: (A), (B), (C)

EM02.10GATE 2022, Q15MCQ | 1 mark

On the surface of a spherical shell enclosing a charge free region, the electrostatic potential values are as follows: One quarter of the area has potential \(\phi_0\), another quarter has potential \(2\phi_0\) and the rest has potential \(4\phi_0\). The potential at the centre of the shell is (You can use a property of the solution of Laplace's equation.)

\(\dfrac{11}{4}\phi_0\)
\(\dfrac{11}{2}\phi_0\)
\(\dfrac{7}{3}\phi_0\)
\(\dfrac{7}{4}\phi_0\)

Show solution

Inside a charge-free region the potential satisfies Laplace's equation. Its expansion is \[ V_{in}(r,\theta)=\sum_{\ell}A_\ell\,r^\ell P_\ell(\cos\theta)=A_0+A_1rP_1(\cos\theta)+\cdots, \] so at \(r=0\) only \(A_0\) survives. Since \(P_0=1\), \[ A_0=\frac{1}{4\pi R^2}\oint V(R)\,da , \] i.e. the potential at the centre equals the average of the potential over the sphere (mean-value property). For a concrete choice, take the zones \(0<\theta<\pi/3\), \(\pi/3<\theta<\pi/2\) and \(\pi/2<\theta<\pi\). Their areas are \(\pi R^2\), \(\pi R^2\) and \(2\pi R^2\), i.e. one quarter, one quarter and one half. Then \[ A_0=\frac12\left[\phi_0\int_0^{\pi/3}\!\sin\theta\,d\theta+2\phi_0\int_{\pi/3}^{\pi/2}\!\sin\theta\,d\theta+4\phi_0\int_{\pi/2}^{\pi}\!\sin\theta\,d\theta\right] =\frac12\left[\frac{\phi_0}{2}+\phi_0+4\phi_0\right]=\frac{11}{4}\phi_0 . \] Equivalently, \(\tfrac14\phi_0+\tfrac14(2\phi_0)+\tfrac12(4\phi_0)=\tfrac{11}{4}\phi_0\). \[\boxed{V(0)=\dfrac{11}{4}\phi_0}\]

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Chapter 2 Boundary-Value Problems and Multipoles Previous Year Questions with Solutions

EM02: Boundary-Value Problems and Multipoles

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