JEE Advanced PYQ Physics – Electric Dipole, Electric Flux and Gauss’s Law Questions

Q 1 :

Two co-axial conducting cylinders of same length $ℓ$ with radii $\sqrt{2} R$ and $2 R$ are kept, as shown in Fig. 1. The charge on the inner cylinder is $Q$ and the outer cylinder is grounded. The annular region between the cylinders is filled with a material of dielectric constant $k = 5$ . Consider an imaginary plane of the same length $ℓ$ at a distance $R$ from the common axis of the cylinders. This plane is parallel to the axis of the cylinders. The cross-sectional view of this arrangement is shown in Fig. 2. Ignoring edge effects, the flux of the electric field through the plane is ( $ε_{0}$ is the permittivity of free space): [2025]

$\frac{Q}{30 ε_{0}}$
$\frac{Q}{15 ε_{0}}$
$\frac{Q}{60 ε_{0}}$
$\frac{Q}{120 ε_{0}}$

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(3)

For symmetry, assume $ℓ$ is very large.

Outside cylinder will have zero electric field, so the flux generated on the plate will be due to the inner cylinder only in sections AB and CD.

And $ϕ_{A B} = ϕ_{C D}$ and $ϕ_{B C} = 0$

Flux through an element will be $d ϕ = \vec{E} \cdot \vec{d S}$

$d ϕ = \frac{2 k λ}{r} d y \cdot ℓ \cdot \cos θ$ ...(i)

From figure

$\cos θ = \frac{R}{r} \Rightarrow r = R \sec θ$

$\tan θ = \frac{y}{R} \Rightarrow y = R \tan θ$ $\Rightarrow d y = R \sec^{2} θ d θ$

$d ϕ = \frac{2 k λ}{R \sec θ} \cdot R \sec^{2} θ \cdot ℓ \cdot \cos θ d θ$

$d ϕ = 2 k λ ℓ d θ$

Therefore, $\int_{0}^{ϕ A B} d ϕ = 2 k λ ℓ \int_{π / 4}^{π / 3} d θ$ $= 2 k λ ℓ [\frac{π}{3} - \frac{π}{4}]$

$ϕ_{A B} = 2 k Q [\frac{π}{12}]$

$ϕ_{A B} = 2 (\frac{1}{4 π ε_{0} ε_{r}}) Q [\frac{π}{12}]$

$ϕ_{A B} = \frac{Q}{120 ε_{0}}$

$ϕ_{plate} = ϕ_{A B} + ϕ_{B C} + ϕ_{C D}$

$ϕ_{plate} = \frac{Q}{120 ε_{0}} + 0 + \frac{Q}{120 ε_{0}} = \frac{Q}{60 ε_{0}}$

Q 2 :

A small electric dipole ${\vec{p}}_{0}$ , having a moment of inertia $I$ about its center, is kept at a distance $r$ from the center of a spherical shell of radius $R$ . The surface charge density $σ$ is uniformly distributed on the spherical shell. The dipole is initially oriented at a small angle $θ$ as shown in the figure. While staying at a distance $r$ , the dipole is free to rotate about its center.

If released from rest, then which of the following statement(s) is(are) correct? [2024]

[ $ε_{0}$ is the permittivity of free space.]

The dipole will undergo small oscillations at any finite value of $r$ .
The dipole will undergo small oscillations at any finite value of $r > R$ .
The dipole will undergo small oscillations with an angular frequency of $\sqrt{\frac{2 σ p_{0}}{ε_{0} I}}$ at $r = 2 R$ .
The dipole will undergo small oscillations with an angular frequency of $\sqrt{\frac{σ p_{0}}{100 ε_{0} I}}$ at $r = 10 R$ .

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Select one or more options

(2, 4)

The electric field inside the sphere is zero, so the dipole will oscillate when $r > R$ .

Hence option (2) is correct and option (1) is incorrect.

For $r > R$ , $E = \frac{σ R^{2}}{ε_{0} r^{2}}$

$ω = \sqrt{\frac{P E}{I}} = \sqrt{\frac{P_{0} σ R^{2}}{I ε_{0} r^{2}}}$

When $r = 2 R;$ $ω = \sqrt{\frac{P_{0} σ R^{2}}{I ε_{0} {(2 R)}^{2}}}$

or, $ω = \sqrt{\frac{P_{0} σ}{4 I ε_{0}}}$

Therefore option (3) is incorrect.

When $r = 10 R;$ $ω = \sqrt{\frac{P_{0} σ R^{2}}{I ε_{0} {(10 R)}^{2}}}$

$∴$ $ω = \sqrt{\frac{P_{0} σ}{100 I ε_{0}}}$

Hence option (4) is correct.

Q 3 :

Consider an electric field $\vec{E} = E_{0} \hat{x}$ where $E_{0}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is [2011]

$2 E_{0} a^{2}$
$\sqrt{2} E_{0} a^{2}$
$E_{0} a^{2}$
$\frac{E_{0} a^{2}}{\sqrt{2}}$

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(3)

Given $\vec{E} = E_{0} \hat{x}$

i.e., the electric field $\vec{E}$ acts along the $+ x$ -direction and is constant.

Therefore, the electric flux through the shaded portion whose area $\vec{A} = a \times \sqrt{2} a = \sqrt{2} a^{2}$ is

$ϕ = \vec{E} \cdot \vec{A}$ $= E A \cos θ$ $= E_{0} (\sqrt{2} a^{2}) \cos 45 °$ $= E_{0} (\sqrt{2} a^{2}) \times (\frac{1}{\sqrt{2}})$ $= E_{0} a^{2}$ $(∵ angle between \vec{E} and \vec{A}, Q = 45 °)$

Q 4 :

A disc of radius $a / 4$ having a uniformly distributed charge 6C is placed in the $x - y$ plane with its centre at $(- a / 2, 0, 0)$ . A rod of length $a$ carrying a uniformly distributed charge 8C is placed on the $x$ -axis from $x = a / 4$ to $x = 5 a / 4$ . Two point charges $- 7 C$ and 3C are placed at $(a / 4, - a / 4, 0)$ and $(- 3 a / 4, 3 a / 4, 0)$ , respectively. Consider a cubical surface formed by six surfaces $x = \pm \frac{a}{2}, y = \pm \frac{a}{2}, z = \pm \frac{a}{2} .$ The electric flux through this cubical surface is [2009]

$- \frac{2 C}{ε_{0}}$
$\frac{2 C}{ε_{0}}$
$\frac{10 C}{ε_{0}}$
$\frac{12 C}{ε_{0}}$

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(1)

From the figure, total charge enclosed in the cubical surface is $q_{in} = 3 C + 2 C - 7 C = - 2 C$

According to Gauss's theorem, the electric flux through the cube is

$ϕ = \frac{q_{in}}{ε_{0}}$ $= \frac{- 2 C}{ε_{0}}$

Q 5 :

A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be [2004]

due to $q_{1}$ and $q_{2}$ only
due to $q_{2}$ only
zero
due to all

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(4)

The flux through the Gaussian surface is due to the charges inside the Gaussian surface. But the electric field on the Gaussian surface is the vector sum of the electric fields due to all the charges, i.e., $q_{1}, - q_{1}$ and $q_{2}$ .

Q 6 :

A charge is kept at the central point $P$ of a cylindrical region. The two edges subtend a half-angle $θ$ at $P$ , as shown in the figure. When $θ = 30 °$ , then the electric flux through the curved surface of the cylinder is $Φ$ . If $θ = 60 °$ , then the electric flux through the curved surface becomes $\frac{Φ}{\sqrt{n}},$ where the value of $n$ is _________. [2024]

(3)

Solid angle subtended at the centre by the plane surface

$Ω = 2 \times 2 π (1 - \cos θ)$

$= 4 π - 4 π \cos θ$

So, solid angle made by the curved surface $= 4 π - Ω$

$= 4 π - (4 π - 4 π \cos θ)$ $= 4 π \cos θ$

Flux through curved surface $= \frac{Q}{ε_{0}} \cos θ$

$ϕ_{30 °} = ϕ = \frac{4 π \cos 30 °}{4 π} \cdot \frac{Q}{ε_{0}} = \cos 30 ° \frac{Q}{ε_{0}}$

$ϕ_{60 °} = \frac{4 π \cos 60 °}{4 π} \cdot \frac{Q}{ε_{0}} = \cos 60 ° \frac{Q}{ε_{0}}$

$\frac{ϕ_{30 °}}{ϕ_{60 °}} = \frac{\cos 30 °}{\cos 60 °} = \sqrt{3}$

$\frac{ϕ}{ϕ_{60 °}} = \sqrt{3}$ $\Rightarrow ϕ_{60 °} = \frac{ϕ}{\sqrt{3}}$

$∴$ $n = 3$

Q 7 :

A charge $q$ is surrounded by a closed surface consisting of an inverted cone of height $h$ and base radius $R$ , and a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n}{6} \frac{q}{ε_{0}}$ (in SI units). The value of $n$ is _______. [2022]

(3)

$ϕ_{cone} = \frac{q}{2 ε_{0}} = \frac{3 q}{6 ε_{0}}$

$∴$ $n = 3$

Q 8 :

An infinitely long uniform line charge distribution of charge per unit length $λ$ lies parallel to the $y$ -axis in the $y - z$ plane at $z = \frac{\sqrt{3}}{2} a$ (see figure). If the magnitude of the flux of the electric field through the rectangular surface $A B C D$ lying in the $x - y$ plane with its centre at the origin is $\frac{λ L}{n ε_{0}}$ ( $ε_{0}$ = permittivity of free space), then the value of $n$ is [2015]

(6)

From the figure $\tan θ = \frac{B C}{O B} = \frac{a / 2}{\sqrt{3} a / 2} = \frac{1}{\sqrt{3}}$

$∴$ $θ = 30 °$

Electric flux through the complete cylinder by Gauss's theorem

$ϕ_{cylinder} = \frac{q_{in}}{ε_{0}} = \frac{λ L}{ε_{0}}$

(where L = length of cylinder)

$∴$ Electric flux passing through the cylindrical surface i.e., for $60^{o}$ angle $= \frac{λ L}{6 ε_{0}}$

Hence, $n = 6$

Q 9 :

A circular disc of radius $R$ carries surface charge density $σ (r) = σ_{0} (1 - \frac{r}{R}),$ where $σ_{0}$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $ϕ_{0}$ . Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $ϕ$ . Then the ratio $\frac{ϕ_{0}}{ϕ}$ is _________. [2020]

(6.40)

Let us consider a ring element of radius $r$ and thickness dr.

Surface charge density of a disc of radius R,

$σ (r) = σ_{0} (1 - \frac{r}{R})$

Charge of disc element,

$d q = σ_{0} (1 - \frac{r}{R}) 2 π r d r$

Now, from Gauss's theorem, electric flux through a large spherical surface that encloses the charged disc completely is

$ϕ_{0} = \frac{\int d q}{ε_{0}} = \frac{\int_{0}^{R} σ_{0} (1 - \frac{r}{R}) 2 π r d r}{ε_{0}}$

Electric flux through another spherical surface of radius $\frac{R}{4}$

$ϕ = \frac{\int d q}{ε_{0}} = \frac{\int_{0}^{R / 4} σ_{0} (1 - \frac{r}{R}) 2 π r d r}{ε_{0}}$

$∴$ $\frac{ϕ_{0}}{ϕ} = \frac{σ_{0} 2 π \int_{0}^{R} (r - \frac{r^{2}}{R}) d r}{σ_{0} 2 π \int_{0}^{R / 4} (r - \frac{r^{2}}{R}) d r}$ $= \frac{[\frac{R^{2}}{2} - \frac{R^{2}}{3}]}{[\frac{R^{2}}{32} - \frac{R^{2}}{3 \times 64}]}$

$= \frac{\frac{R^{2}}{6}}{\frac{5 R^{2}}{192}} = \frac{32}{5} = 6.40$

Q 10 :

A charged shell of radius $R$ carries a total charge $Q$ . Given $ϕ$ as the flux of electric field through a closed cylindrical surface of height $h$ , radius $r$ and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct? [2019]

[ $ε_{0}$ is the permittivity of free space]

If $h > 2 R$ and $r = \frac{3 R}{5}$ , then $ϕ = \frac{Q}{5 ε_{0}}$
If $h > 2 R$ and $r > R$ , then $ϕ = \frac{Q}{ε_{0}}$
If $h < \frac{8 R}{5}$ and $r = \frac{3 R}{5}$ , then $ϕ = 0$
If $h > 2 R$ and $r > \frac{4 R}{5}$ , then $ϕ = \frac{Q}{5 ε_{0}}$

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Select one or more options

(1, 2, 3)

(1) $For h > 2 R and r = \frac{3 R}{5}$

$\sin θ = \frac{3 R / 5}{R} = \frac{3}{5}$ $= 37 °$

$q_{in} = Q (1 - \cos 37 °) = Q [1 - \frac{4}{5}] = \frac{Q}{5}$

$From Gauss's theorem$ $Q = \frac{q_{in}}{ε_{0}}$

$∴$ $ϕ = \frac{Q}{5 ε_{0}}$

(2) $For h > 2 R and r > R$

$ϕ = \frac{q_{in}}{ε_{0}} = \frac{Q}{ε_{0}}$

(3) $For h < \frac{8}{5} R and r = \frac{3}{5} R$

$ϕ = \frac{q_{in}}{ε_{0}} = 0$

(4) $For h > 2 R and r > \frac{4}{5} R$

$\sin θ = \frac{4 R / 5}{R} = \frac{4}{5} = 0.8$ $\Rightarrow$ $θ = 53 °$

$q_{in} = Q (1 - \cos θ) = Q (1 - \frac{3}{5}) = \frac{2 Q}{5}$

$∴$ $ϕ = \frac{q_{in}}{ε_{0}} = \frac{2 Q}{5 ε_{0}}$

Q 11 :

An infinitely long thin non-conducting wire is parallel to the $z$ -axis and carries a uniform line charge density $λ$ . It pierces a thin non-conducting spherical shell of radius $R$ in such a way that the arc $P Q$ subtends an angle $120 °$ at the centre $O$ of the spherical shell, as shown in the figure. The permittivity of free space is $ε_{0}$ . Which of the following statements is (are) true? [2018]

The electric flux through the shell is $\frac{\sqrt{3} R λ}{ε_{0}}$
The $z$ -component of the electric field is zero at all the points on the surface of the shell.
The electric flux through the shell is $\frac{\sqrt{2} R λ}{ε_{0}}$
The electric field is normal to the surface of the shell at all points.

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Select one or more options

(1, 2)

According to Gauss's law, electric flux

$ϕ = \frac{1}{ε_{0}} q_{in} = \frac{1}{ε_{0}} [λ \times 2 R \sin 60 °]$

$= \frac{\sqrt{3} λ R}{ε_{0}}$

$A B = R \sin 60 °$ or $A C = 2 R \sin 60 °$

Also, the electric field is perpendicular to the wire therefore its z-component is zero.

Q 12 :

A point charge $+ Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct? [2017]

The electric flux passing through the curved surface of the hemisphere is $- \frac{Q}{2 ε_{0}} (1 - \frac{1}{\sqrt{2}})$
Total flux through the curved and the flat surfaces is $\frac{Q}{ε_{0}}$
The component of the electric field normal to the flat surface is constant over the surface.
The circumference of the flat surface is an equipotential.

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Select one or more options

(1, 4)

The circumference of the flat surface is an equipotential $V = \frac{K Q}{\sqrt{2} R}$

because the circumference is equidistant from $+ Q$ . The component of electric field perpendicular to the flat surface is $E \cos θ .$

Here E as well as $θ$ changes for different points on the flat surface. The total flux through the curved and flat surfaces should be less than $\frac{Q}{ε_{0}} .$

The solid angle subtended by the flat surface at $P$

$= 2 π (1 - \cos θ) = 2 π (1 - \cos 45 °)$ $= 2 π (1 - \frac{1}{\sqrt{2}})$

$∴$ Flux passing through the curved surface

$= - \frac{Q}{ε_{0}} \cdot \frac{2 π (1 - \frac{1}{\sqrt{2}})}{4 π}$ $= - \frac{Q}{2 ε_{0}} (1 - \frac{1}{\sqrt{2}})$

Q 13 :

A cubical region of side $a$ has its centre at the origin. It encloses three fixed point charges, $- q$ at $(0, - a / 4, 0)$ , $+ 3 q$ at $(0, 0, 0)$ and $- q$ at $(0, + a / 4, 0)$ . Choose the correct option(s). [2012]

The net electric flux crossing the plane $x = + a / 2$ is equal to the net electric flux crossing the plane $x = - a / 2$ .
The net electric flux crossing the plane $y = + a / 2$ is more than the net electric flux crossing the plane $y = - a / 2$ .
The net electric flux crossing the entire region is $\frac{q}{ε_{0}} .$
The net electric flux crossing the plane $z = + a / 2$ is equal to the net electric flux crossing the plane $x = + a / 2$

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Select one or more options

(1, 3, 4)

Due to symmetry, the net electric flux passing through

$x = + \frac{a}{2}, x = - \frac{a}{2}, z = + \frac{a}{2}$ is the same.

According to Gauss's theorem, the net electric flux crossing through any closed surface $ϕ = \frac{q_{in}}{ε_{0}}$

$= \frac{- q + 3 q - q}{ε_{0}} = \frac{q}{ε_{0}}$

Q 14 :

List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude $p$ , oriented as marked by arrows in the figures. In all the configurations the dipoles are fixed such that they are at a distance $2 r$ apart along the $x$ direction. The midpoint of the line joining the two dipoles is $X$ . The possible resultant electric fields $\vec{E}$ at $X$ are given in List-II. Choose the option that describes the correct match between the entries in List-I to those in List-II. [2025]

List-I List-II

(P) (1) $\vec{E} = 0$

(Q) (2) $\vec{E} = \frac{p}{2 π ε_{0} r^{3}} \hat{j}$

(R) (3) $\vec{E} = - \frac{p}{4 π ε_{0} r^{3}} (\hat{i} - \hat{j})$

(S) (4) $\vec{E} = \frac{p}{4 π ε_{0} r^{3}} (2 \hat{i} - \hat{j})$

(5) $\vec{E} = \frac{p}{π ε_{0} r^{3}} \hat{i}$

(P) → (3), (Q) → (1), (R) → (2), (S) → (4)
(P) → (4), (Q) → (5), (R) → (3), (S) → (1)
(P) → (2), (Q) → (1), (R) → (4), (S) → (5)
(P) → (2), (Q) → (1), (R) → (3), (S) → (5)

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(3)

(P) ${\vec{E}}_{net} = \frac{- 2 k P}{r^{3}} \hat{j} = \frac{- P \hat{j}}{2 π ε_{0} r^{3}} \to (2)$

(Q) ${\vec{E}}_{net} = 0 \to (1)$

(R) ${\vec{E}}_{net} = \frac{2 P \hat{i}}{4 π ε_{0} r^{3}} - \frac{P \hat{j}}{4 π ε_{0} r^{3}}$

$= \frac{P}{4 π ε_{0} r^{3}} (2 \hat{i} - \hat{j}) \to (4)$

(S)

${\vec{E}}_{net} = \frac{4 k P \hat{i}}{r^{3}} = \frac{4 P \hat{i}}{4 π ε_{0} r^{3}} = \frac{P}{π ε_{0} r^{3}} \hat{i} \to (5)$

Topic Question Set

Q 3 :

Consider an electric field $\vec{E} = E_{0} \hat{x}$ where $E_{0}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is [2011]

Q 5 :

A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be [2004]

Q 12 :

A point charge $+ Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct? [2017]

Q 13 :

A cubical region of side $a$ has its centre at the origin. It encloses three fixed point charges, $- q$ at $(0, - a / 4, 0)$ , $+ 3 q$ at $(0, 0, 0)$ and $- q$ at $(0, + a / 4, 0)$ . Choose the correct option(s). [2012]

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	List-I		List-II
(P)		(1)	$\vec{E} = 0$
(Q)		(2)	$\vec{E} = \frac{p}{2 π ε_{0} r^{3}} \hat{j}$
(R)		(3)	$\vec{E} = - \frac{p}{4 π ε_{0} r^{3}} (\hat{i} - \hat{j})$
(S)		(4)	$\vec{E} = \frac{p}{4 π ε_{0} r^{3}} (2 \hat{i} - \hat{j})$
		(5)	$\vec{E} = \frac{p}{π ε_{0} r^{3}} \hat{i}$

Topic Question Set

Q 3 : Consider an electric field E→=E0x^ where E0 is a constant. The flux through the shaded area (as shown in the figure) due to this field is [2011]

Q 5 : A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be [2004]

Q 7 : A charge q is surrounded by a closed surface consisting of an inverted cone of height h and base radius R, and a hemisphere of radius R as shown in the figure. The electric flux through the conical surface is n6qε0 (in SI units). The value of n is _______. [2022]

Q 12 : A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct? [2017]

Q 13 : A cubical region of side a has its centre at the origin. It encloses three fixed point charges, -q at (0,-a/4,0), +3q at (0,0,0) and -q at (0,+a/4,0). Choose the correct option(s). [2012]

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Q 3 :

Consider an electric field $\vec{E} = E_{0} \hat{x}$ where $E_{0}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is [2011]

Q 5 :

A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be [2004]

Q 12 :

A point charge $+ Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct? [2017]

Q 13 :

A cubical region of side $a$ has its centre at the origin. It encloses three fixed point charges, $- q$ at $(0, - a / 4, 0)$ , $+ 3 q$ at $(0, 0, 0)$ and $- q$ at $(0, + a / 4, 0)$ . Choose the correct option(s). [2012]