Gauss Law and Divergence Questions and Answers

Q: An infinitely large non-conducting thin sheet in the \(xy\) plane (\(z\)=0) carries a uniform surface charge density \(\sigma = 17.70 \times 10^{-12}\) C.m−2. The electric field in the region \(z \lt 0\) is \(\vec{E}_{2} = \hat{x} + 2\hat{y} + 3\hat{z}\). Then, the electric field \(\vec{E}_{1}\) in the region \(z \gt 0\) will be(\(\epsilon_{0} = 8.85 \times 10^{-12}\) C2.N−1.m−2)

The field above the sheet is \(\vec{E}_{1} = \hat{x} + 2\hat{y} + 5\hat{z}\).At a charged surface the tangential component of \(\vec{E}\) is continuous, and the normal component jumps by \(\sigma/\epsilon_{0}\):\[ \begin{aligned} E_{1z} - E_{2z} &= \frac{\sigma}{\epsilon_{0}} = \frac{17.70 \times 10^{-12}}{8.85 \times 10^{-12}} = 2 \\ E_{1z} &= 3 + 2 = 5 \end{aligned} \]The x and y components, 1 and 2, carry over unchanged. Option with \(4\hat{z}\) adds only \(\sigma/2\epsilon_{0}\), the field of the sheet alone on one side; \(3\hat{z}\) ignores the charge, and the last option changes the tangential components, which a surface charge cannot do.

Q: Three infinite plane sheets carrying uniform charge densities \(-\sigma, 2\sigma, 3\sigma\) are placed parallel to the x-z plane at \(y = a, 3a, 4a\), respectively. The electric field at the point \((0, 2a, 0)\) is

The field is \(-\frac{3\sigma}{\varepsilon_{0}}\hat{\jmath}\).An infinite sheet of charge density \(\sigma\) gives a field \(\frac{\sigma}{2\varepsilon_{0}}\) pointing away from it (towards it for negative charge). The point \(y = 2a\) lies above the sheet at \(y = a\) and below the sheets at \(3a\) and \(4a\):\(-\sigma\) at \(y = a\): field points towards the sheet, \(-\frac{\sigma}{2\varepsilon_{0}}\hat{\jmath}\).\(2\sigma\) at \(y = 3a\): points away, downwards, \(-\frac{2\sigma}{2\varepsilon_{0}}\hat{\jmath}\).\(3\sigma\) at \(y = 4a\): points away, downwards, \(-\frac{3\sigma}{2\varepsilon_{0}}\hat{\jmath}\).\[ \begin{aligned} \vec{E} &= -\frac{(1 + 2 + 3)\sigma}{2\varepsilon_{0}}\hat{\jmath} \\ &= -\frac{3\sigma}{\varepsilon_{0}}\hat{\jmath} \end{aligned} \]Getting the sign of one sheet wrong gives \(-\frac{2\sigma}{\varepsilon_{0}}\) or \(\frac{\sigma}{\varepsilon_{0}}\).

Q: Given a spherically symmetric charge density \(\rho(r) = \begin{cases} kr^{2}, & r \lt R \\ 0, & r \gt R \end{cases}\) (k being a constant), the electric field for \(r \lt R\) is (take the total charge as Q)

The field inside is \(\frac{Qr^{3}}{4\pi\varepsilon_{0}R^{5}}\hat{r}\).First express k through the total charge, then apply Gauss's law to a sphere of radius \(r \lt R\):\[ \begin{aligned} Q &= \int_{0}^{R} kr^{2}\,4\pi r^{2}dr = \frac{4\pi k R^{5}}{5} \\ q(r) &= \frac{4\pi k r^{5}}{5} = Q\frac{r^{5}}{R^{5}} \\ E &= \frac{q(r)}{4\pi\varepsilon_{0}r^{2}} = \frac{Qr^{3}}{4\pi\varepsilon_{0}R^{5}} \end{aligned} \]The last option is the field outside the sphere. The other two have the wrong power of r or the wrong numerical factor; at \(r = R\) only the keyed answer matches the outside field \(\frac{Q}{4\pi\varepsilon_{0}R^{2}}\).

Q: Given below are the two statements :Statement I : Gauss's law can also be applied to non-Gaussian surfaces.Statement II : Gauss's law is useful to find the electric field from the knowledge of enclosed charge and surface.In the light of the above statements, choose the most appropriate answer from the options given below.

The answer is Statement I is false but Statement II is true.Statement I is false: Gauss's law states that the total electric flux through a closed surface equals the enclosed charge divided by \(\varepsilon_{0}\). The closed surface used is by definition a Gaussian surface, so the law is not applied to anything else, and it says nothing about open surfaces.Statement II is true: when the charge distribution is symmetric, choosing a Gaussian surface over which the field is constant lets the field be found directly from the enclosed charge, as for a point charge, a long line charge or a charged sphere.

Q: A hollow sphere of a charge does not produce an electric field at any __________.

A charged hollow sphere produces no electric field at any interior point.On a hollow conducting sphere the charge sits on the outer surface. Apply Gauss's law to any spherical surface drawn inside the sphere: it encloses no charge, so the flux through it, and by symmetry the field at every point on it, is zero. The potential inside is constant and equal to that at the surface.Outside the sphere the field is the same as that of a point charge at the centre, E = kQ/r2. It weakens with distance but is never zero at any outer point, whether 2 m or 10 m away.

Q: Uniformly charged solid sphere of radius R has a volume charge density ρ. The strength of electric field, at a distance r, (\(r \gt R\)), from its centre is :

The field is \(\frac{\rho R^{3}}{3\epsilon_{0} r^{2}}\).Outside a spherically symmetric charge distribution the field is that of a point charge equal to the total charge. The total charge is \(Q = \frac{4}{3}\pi R^{3}\rho\). By Gauss's law on a sphere of radius \(r \gt R\):\[ \begin{aligned} E \cdot 4\pi r^{2} &= \frac{Q}{\epsilon_{0}} = \frac{4\pi R^{3}\rho}{3\epsilon_{0}} \\ E &= \frac{\rho R^{3}}{3\epsilon_{0} r^{2}} \end{aligned} \]The field must fall off as \(1/r^{2}\) outside, which rules out the options growing as \(r^{3}\) or falling only as \(1/r\). The expression \(\rho r/3\epsilon_{0}\) (not offered) would be the field inside the sphere.

Q: A point charge q is placed at a corner of a cube of side-length l. The electric flux through one of the cube faces not passing through the charge q is :

The flux is \( \frac{q}{24\epsilon_{0}} \).A charge at a corner is shared by 8 cubes that together surround it, so by symmetry the given cube receives \( \frac{q}{8\epsilon_{0}} \) in total. The three faces that meet at the corner contain the charge in their planes, so the field lies along them and their flux is zero. The other three faces share the flux equally:\[ \begin{aligned} \Phi &= \frac{1}{3} \times \frac{q}{8\epsilon_{0}} \\ &= \frac{q}{24\epsilon_{0}} \end{aligned} \]\( \frac{q}{6\epsilon_{0}} \) is the flux through one face when the charge is at the centre of the cube, and \( \frac{q}{3\epsilon_{0}} \) and \( \frac{ql^{2}}{\epsilon_{0}} \) do not follow from Gauss's law.

Q: A rectangle of cross-sectional area 'A' is placed in a uniform constant electric field \(\overline{E}\). The plane of the rectangle makes an angle of 30° with the direction of the electric field. The electric flux through the rectangle is :

The flux is AE/2.Flux is \(\Phi = EA\cos\theta\), where \(\theta\) is the angle between the field and the normal to the surface. The plane makes 30° with the field, so the normal makes 90° - 30° = 60° with it:\[ \Phi = EA\cos 60^{\circ} = \frac{AE}{2} \]The option \(AE\sqrt{3}/2\) is the common slip of putting 30° into the cosine, treating it as the angle with the normal. AE would need the field perpendicular to the plane, and zero would need it parallel.

Q: A solid sphere of radius R with its center at origin has a volume charge density \(\rho = A\sqrt{r}\), where A is a constant. The magnitude of electric field, \(\bar{E}\), at \(r \lt R\), varies with r as ........ .

Inside the sphere, \(E \propto r^{3/2}\).By Gauss's law, the field at radius r depends on the charge enclosed:\[ \begin{aligned} q(r) &= \int_{0}^{r} A\sqrt{r'}\, 4\pi r'^{2}\, dr' \\ &= \frac{8\pi A}{7} r^{7/2} \\ E &= \frac{q(r)}{4\pi\epsilon_{0}r^{2}} = \frac{2A}{7\epsilon_{0}} r^{3/2} \end{aligned} \]For a uniform density the same steps give \(E \propto r\); the extra factor \(\sqrt{r}\) in the density raises it to \(r^{3/2}\). \(r^{-2}\) is the law outside the sphere, and \(r^{\pm 1/2}\) would come from forgetting the \(r^{2}\) of the volume element.

Q: For uniformly charged solid sphere of radius R, which of the following statements is correct about the magnitude of electric field at a distance r from the center of the sphere ?

Inside the sphere the field is proportional to r.Apply Gauss's law to a sphere of radius r < R. It encloses the fraction \((r/R)^{3}\) of the total charge Q:\[ \begin{aligned} E\cdot 4\pi r^{2} &= \frac{Q}{\varepsilon_{0}}\frac{r^{3}}{R^{3}} \\ E &= \frac{Qr}{4\pi\varepsilon_{0}R^{3}} \end{aligned} \]so E grows linearly from zero at the centre to its maximum at the surface, then falls as \(1/r^{2}\) outside. E = 0 inside describes a conductor or a charged spherical shell, not a uniformly charged solid sphere. A constant non-zero field or a 1/r field inside does not satisfy Gauss's law for a uniform volume charge.

An infinitely large non-conducting thin sheet in the \(xy\) plane (\(z\)=0) carries a uniform surface charge density \(\sigma = 17.70 \times 10^{-12}\) C.m⁻². The electric field in the region \(z \lt 0\) is \(\vec{E}_{2} = \hat{x} + 2\hat{y} + 3\hat{z}\). Then, the electric field \(\vec{E}_{1}\) in the region \(z \gt 0\) will be

(\(\epsilon_{0} = 8.85 \times 10^{-12}\) C².N⁻¹.m⁻²)

A.\(\vec{E}_{1} = \hat{x} + 2\hat{y} + 5\hat{z}\)
B.\(\vec{E}_{1} = \hat{x} + 2\hat{y} + 4\hat{z}\)
C.\(\vec{E}_{1} = \hat{x} + 2\hat{y} + 3\hat{z}\)
D.\(\vec{E}_{1} = \hat{x} + 4\hat{y} + \hat{z}\)

Exam Year(s)

GATE Physics - 2026,

Answer: A

The field above the sheet is \(\vec{E}_{1} = \hat{x} + 2\hat{y} + 5\hat{z}\).

At a charged surface the tangential component of \(\vec{E}\) is continuous, and the normal component jumps by \(\sigma/\epsilon_{0}\):

\[ \begin{aligned} E_{1z} - E_{2z} &= \frac{\sigma}{\epsilon_{0}} = \frac{17.70 \times 10^{-12}}{8.85 \times 10^{-12}} = 2 \\ E_{1z} &= 3 + 2 = 5 \end{aligned} \]

The x and y components, 1 and 2, carry over unchanged. Option with \(4\hat{z}\) adds only \(\sigma/2\epsilon_{0}\), the field of the sheet alone on one side; \(3\hat{z}\) ignores the charge, and the last option changes the tangential components, which a surface charge cannot do.

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Three infinite plane sheets carrying uniform charge densities \(-\sigma, 2\sigma, 3\sigma\) are placed parallel to the x-z plane at \(y = a, 3a, 4a\), respectively. The electric field at the point \((0, 2a, 0)\) is

A.\(\frac{4\sigma}{\varepsilon_{0}}\hat{\jmath}\)
B.\(-\frac{3\sigma}{\varepsilon_{0}}\hat{\jmath}\)
C.\(-\frac{2\sigma}{\varepsilon_{0}}\hat{\jmath}\)
D.\(\frac{\sigma}{\varepsilon_{0}}\hat{\jmath}\)

Exam Year(s)

IIT JAM - 2018,

Answer: B

The field is \(-\frac{3\sigma}{\varepsilon_{0}}\hat{\jmath}\).

An infinite sheet of charge density \(\sigma\) gives a field \(\frac{\sigma}{2\varepsilon_{0}}\) pointing away from it (towards it for negative charge). The point \(y = 2a\) lies above the sheet at \(y = a\) and below the sheets at \(3a\) and \(4a\):

\(-\sigma\) at \(y = a\): field points towards the sheet, \(-\frac{\sigma}{2\varepsilon_{0}}\hat{\jmath}\).
\(2\sigma\) at \(y = 3a\): points away, downwards, \(-\frac{2\sigma}{2\varepsilon_{0}}\hat{\jmath}\).
\(3\sigma\) at \(y = 4a\): points away, downwards, \(-\frac{3\sigma}{2\varepsilon_{0}}\hat{\jmath}\).

\[ \begin{aligned} \vec{E} &= -\frac{(1 + 2 + 3)\sigma}{2\varepsilon_{0}}\hat{\jmath} \\ &= -\frac{3\sigma}{\varepsilon_{0}}\hat{\jmath} \end{aligned} \]

Getting the sign of one sheet wrong gives \(-\frac{2\sigma}{\varepsilon_{0}}\) or \(\frac{\sigma}{\varepsilon_{0}}\).

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Given a spherically symmetric charge density \(\rho(r) = \begin{cases} kr^{2}, & r \lt R \\ 0, & r \gt R \end{cases}\) (k being a constant), the electric field for \(r \lt R\) is (take the total charge as Q)

A.\(\frac{Qr^{3}}{4\pi\varepsilon_{0}R^{5}}\hat{r}\)
B.\(\frac{3Qr^{2}}{4\pi\varepsilon_{0}R^{4}}\hat{r}\)
C.\(\frac{5Qr^{3}}{8\pi\varepsilon_{0}R^{5}}\hat{r}\)
D.\(\frac{Q}{4\pi\varepsilon_{0}r^{2}}\hat{r}\)

Exam Year(s)

IIT JAM - 2018,

Answer: A

The field inside is \(\frac{Qr^{3}}{4\pi\varepsilon_{0}R^{5}}\hat{r}\).

First express k through the total charge, then apply Gauss's law to a sphere of radius \(r \lt R\):

\[ \begin{aligned} Q &= \int_{0}^{R} kr^{2}\,4\pi r^{2}dr = \frac{4\pi k R^{5}}{5} \\ q(r) &= \frac{4\pi k r^{5}}{5} = Q\frac{r^{5}}{R^{5}} \\ E &= \frac{q(r)}{4\pi\varepsilon_{0}r^{2}} = \frac{Qr^{3}}{4\pi\varepsilon_{0}R^{5}} \end{aligned} \]

The last option is the field outside the sphere. The other two have the wrong power of r or the wrong numerical factor; at \(r = R\) only the keyed answer matches the outside field \(\frac{Q}{4\pi\varepsilon_{0}R^{2}}\).

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Given below are the two statements :

Statement I : Gauss's law can also be applied to non-Gaussian surfaces.
Statement II : Gauss's law is useful to find the electric field from the knowledge of enclosed charge and surface.

In the light of the above statements, choose the most appropriate answer from the options given below.

A.Both Statement I and Statement II are false.
B.Statement I is true and Statement II is false.
C.Statement I is false but Statement II is true.
D.Both Statement I and Statement II are true.

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: C

The answer is Statement I is false but Statement II is true.

Statement I is false: Gauss's law states that the total electric flux through a closed surface equals the enclosed charge divided by \(\varepsilon_{0}\). The closed surface used is by definition a Gaussian surface, so the law is not applied to anything else, and it says nothing about open surfaces.

Statement II is true: when the charge distribution is symmetric, choosing a Gaussian surface over which the field is constant lets the field be found directly from the enclosed charge, as for a point charge, a long line charge or a charged sphere.

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A hollow sphere of a charge does not produce an electric field at any __________.

A.point beyond 10 meters
B.outer point
C.interior point
D.point beyond 2 meters

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: C

A charged hollow sphere produces no electric field at any interior point.

On a hollow conducting sphere the charge sits on the outer surface. Apply Gauss's law to any spherical surface drawn inside the sphere: it encloses no charge, so the flux through it, and by symmetry the field at every point on it, is zero. The potential inside is constant and equal to that at the surface.

Outside the sphere the field is the same as that of a point charge at the centre, E = kQ/r². It weakens with distance but is never zero at any outer point, whether 2 m or 10 m away.

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Uniformly charged solid sphere of radius R has a volume charge density ρ. The strength of electric field, at a distance r, (\(r \gt R\)), from its centre is :

A.\(\frac{\rho r^{3}}{3 \epsilon_{0} R^{2}}\)
B.\(\frac{\rho R^{3}}{3 \epsilon_{0} r^{2}}\)
C.\(\frac{\rho R^{3}}{3 \epsilon_{0} r}\)
D.\(\frac{\rho r^{3}}{3 \epsilon_{0} R}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The field is \(\frac{\rho R^{3}}{3\epsilon_{0} r^{2}}\).

Outside a spherically symmetric charge distribution the field is that of a point charge equal to the total charge. The total charge is \(Q = \frac{4}{3}\pi R^{3}\rho\). By Gauss's law on a sphere of radius \(r \gt R\):

\[ \begin{aligned} E \cdot 4\pi r^{2} &= \frac{Q}{\epsilon_{0}} = \frac{4\pi R^{3}\rho}{3\epsilon_{0}} \\ E &= \frac{\rho R^{3}}{3\epsilon_{0} r^{2}} \end{aligned} \]

The field must fall off as \(1/r^{2}\) outside, which rules out the options growing as \(r^{3}\) or falling only as \(1/r\). The expression \(\rho r/3\epsilon_{0}\) (not offered) would be the field inside the sphere.

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A point charge q is placed at a corner of a cube of side-length l. The electric flux through one of the cube faces not passing through the charge q is :

A.\( \frac{ql^{2}}{\epsilon_{0}} \)
B.\( \frac{q}{3\epsilon_{0}} \)
C.\( \frac{q}{6\epsilon_{0}} \)
D.\( \frac{q}{24\epsilon_{0}} \)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The flux is \( \frac{q}{24\epsilon_{0}} \).

A charge at a corner is shared by 8 cubes that together surround it, so by symmetry the given cube receives \( \frac{q}{8\epsilon_{0}} \) in total. The three faces that meet at the corner contain the charge in their planes, so the field lies along them and their flux is zero. The other three faces share the flux equally:

\[ \begin{aligned} \Phi &= \frac{1}{3} \times \frac{q}{8\epsilon_{0}} \\ &= \frac{q}{24\epsilon_{0}} \end{aligned} \]

\( \frac{q}{6\epsilon_{0}} \) is the flux through one face when the charge is at the centre of the cube, and \( \frac{q}{3\epsilon_{0}} \) and \( \frac{ql^{2}}{\epsilon_{0}} \) do not follow from Gauss's law.

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A rectangle of cross-sectional area 'A' is placed in a uniform constant electric field \(\overline{E}\). The plane of the rectangle makes an angle of 30° with the direction of the electric field. The electric flux through the rectangle is :

A.AE
B.\(\frac{AE}{2}\)
C.\(AE\frac{\sqrt{3}}{2}\)
D.Zero

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The flux is AE/2.

Flux is \(\Phi = EA\cos\theta\), where \(\theta\) is the angle between the field and the normal to the surface. The plane makes 30° with the field, so the normal makes 90° - 30° = 60° with it:

\[ \Phi = EA\cos 60^{\circ} = \frac{AE}{2} \]

The option \(AE\sqrt{3}/2\) is the common slip of putting 30° into the cosine, treating it as the angle with the normal. AE would need the field perpendicular to the plane, and zero would need it parallel.

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A solid sphere of radius R with its center at origin has a volume charge density \(\rho = A\sqrt{r}\), where A is a constant. The magnitude of electric field, \(\bar{E}\), at \(r \lt R\), varies with r as ........ .

A.\(E \propto r^{-\frac{1}{2}}\)
B.\(E \propto r^{\frac{1}{2}}\)
C.\(E \propto r^{\frac{3}{2}}\)
D.\(E \propto r^{-2}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

Inside the sphere, \(E \propto r^{3/2}\).

By Gauss's law, the field at radius r depends on the charge enclosed:

\[ \begin{aligned} q(r) &= \int_{0}^{r} A\sqrt{r'}\, 4\pi r'^{2}\, dr' \\ &= \frac{8\pi A}{7} r^{7/2} \\ E &= \frac{q(r)}{4\pi\epsilon_{0}r^{2}} = \frac{2A}{7\epsilon_{0}} r^{3/2} \end{aligned} \]

For a uniform density the same steps give \(E \propto r\); the extra factor \(\sqrt{r}\) in the density raises it to \(r^{3/2}\). \(r^{-2}\) is the law outside the sphere, and \(r^{\pm 1/2}\) would come from forgetting the \(r^{2}\) of the volume element.

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10.

For uniformly charged solid sphere of radius R, which of the following statements is correct about the magnitude of electric field at a distance r from the center of the sphere ?

A.E = 0 for \(r \lt R\)
B.E = non-zero constant for \(r \lt R\)
C.E proportional to r, for \(r \lt R\)
D.E proportional to 1/r, for \(r \lt R\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

Inside the sphere the field is proportional to r.

Apply Gauss's law to a sphere of radius r < R. It encloses the fraction \((r/R)^{3}\) of the total charge Q:

\[ \begin{aligned} E\cdot 4\pi r^{2} &= \frac{Q}{\varepsilon_{0}}\frac{r^{3}}{R^{3}} \\ E &= \frac{Qr}{4\pi\varepsilon_{0}R^{3}} \end{aligned} \]

so E grows linearly from zero at the centre to its maximum at the surface, then falls as \(1/r^{2}\) outside. E = 0 inside describes a conductor or a charged spherical shell, not a uniformly charged solid sphere. A constant non-zero field or a 1/r field inside does not satisfy Gauss's law for a uniform volume charge.

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