Q. No.A hollow conducting sphere of radius \(1~\text{m}\) is given a positive charge of \(10~\mu\text{C}\). The electric field at the centre of the hollow sphere will be:
| 1. | \(60\times10^{3}~\text{Vm}^{-1}\) | 2. | \(90\times10^{3}~\text{Vm}^{-1}\) |
| 3. | zero | 4. | infinite |
Subtopic: Gauss's Law |
93%
Level 1: 80%+
AIPMT - 1998
Hints
Links
A sphere encloses an electric dipole with charges \(\pm 3\times 10^{-6}~\text{C}.\) What is the total electric flux through the sphere?
1. \(-3\times 10^{-6}~\text{Nm}^2/\text{C}\)
2. zero
3. \(3\times 10^{-6}~\text{Nm}^2/\text{C}\)
4. \(6\times 10^{-6}~\text{Nm}^2/\text{C}\)
1. \(-3\times 10^{-6}~\text{Nm}^2/\text{C}\)
2. zero
3. \(3\times 10^{-6}~\text{Nm}^2/\text{C}\)
4. \(6\times 10^{-6}~\text{Nm}^2/\text{C}\)
Subtopic: Gauss's Law |
92%
Level 1: 80%+
Please attempt this question first.
Hints
Links
Please attempt this question first.
Five charges \(+q,+5 q,-2 q,+3 q\) and \(-4 q\) are situated a shown in the figure. The electric flux due to this configuration through the surface S is
1. \(\frac{4 q}{\epsilon_0}\)
2. \(\frac{q}{\epsilon_0}\)
3. \(\frac{5 q}{\epsilon_0}\)
4. \(\frac{3 q}{\epsilon_0}\)
1. \(\frac{4 q}{\epsilon_0}\)
2. \(\frac{q}{\epsilon_0}\)
3. \(\frac{5 q}{\epsilon_0}\)
4. \(\frac{3 q}{\epsilon_0}\)
Subtopic: Gauss's Law |
91%
Level 1: 80%+
Please attempt this question first.
Hints
Please attempt this question first.
The electric flux through the surface:
| 1. | in figure-(iv) is the largest |
| 2. | in figure-(iii) is the least |
| 3. | in figure-(ii) is same as figure-(iii) but is smaller than figure-(iv) |
| 4. | is the same for all the figures |
Subtopic: Gauss's Law |
91%
Level 1: 80%+
Hints
The following figure shows some point charges placed and a closed surface \(S\). Total electric flux linked with the closed surface \(S\) is:
1. \(\dfrac{-3q}{\varepsilon_0}\)
2. \(\dfrac{8q}{\varepsilon_0}\)
3. \(\dfrac{4q}{\varepsilon_0}\)
4. \(\dfrac{-4q}{\varepsilon_0}\)
1. \(\dfrac{-3q}{\varepsilon_0}\)
2. \(\dfrac{8q}{\varepsilon_0}\)
3. \(\dfrac{4q}{\varepsilon_0}\)
4. \(\dfrac{-4q}{\varepsilon_0}\)
Subtopic: Gauss's Law |
91%
Level 1: 80%+
Please attempt this question first.
Hints
Please attempt this question first.
The electric field in a region is given by \(\vec{E}=(6 \hat{i}+7 \hat{j}+8 \hat{k })~\text{V/m}.\) If a surface of area \(30~\text{m}^2\) lies in the \(yz \text-\)plane, what is the electric flux through this surface (in V-m)?
| 1. | \(180\) | 2. | \(150\) |
| 3. | \(170\) | 4. | \(130\) |
Subtopic: Gauss's Law |
89%
Level 1: 80%+
Please attempt this question first.
Hints
Please attempt this question first.
The electric field in a region is given \(\vec{E}=\dfrac{4}{5}\hat{j}~\text{N/C}\). The electric flux (in SI units) through the rectangular surface of area \(5~\text{m}^2\) (parallel to \(XZ\)-plane) is:
1. \(1\)
2. \(2\)
3. \(3\)
4. \(4\)
Subtopic: Gauss's Law |
88%
Level 1: 80%+
Please attempt this question first.
Hints
Links
Please attempt this question first.
An electric dipole is enclosed by a Gaussian surface (see figure). The total electric flux through the surface is:
| 1. | \(q / \varepsilon_0\) | 2. | zero |
| 3. | \(-q / \varepsilon_0\) | 4. | \(2q / \varepsilon_0\) |
Subtopic: Gauss's Law |
88%
Level 1: 80%+
Please attempt this question first.
Hints
Please attempt this question first.
Given below are two statements:
| Assertion (A): | When an electric dipole is completely enclosed by a closed Gaussian surface, the total electric flux through the surface is zero. |
| Reason (R): | The net charge enclosed within the surface is zero. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | Both (A) and (R) are False. |
Subtopic: Gauss's Law |
87%
Level 1: 80%+
Please attempt this question first.
Hints
Please attempt this question first.
If there were only one type of charge in the universe, then:
| 1. | \(\oint_{s} \vec{E} \cdot d \vec{s} \neq 0\) on any surface. |
| 2. | \(\oint_{s} \vec{E} \cdot d \vec{s}=0\) if the charge is outside the surface. |
| 3. | \(\oint_{s} \vec{E} \cdot d \vec{s}=\frac{q}{\varepsilon_{0}}\) if charges of magnitude \(q\) were inside the surface. |
| 4. | Both (2) and (3) are correct. |
Subtopic: Gauss's Law |
87%
Level 1: 80%+
Hints
© 2026 GoodEd Technologies Pvt. Ltd.