Electric field at a point varies as r⁰ for

(1) An electric dipole

(2) A point charge

(3) A plane infinite sheet of charge

(4) A line charge of infinite length

Total electric flux coming out of a unit positive charge put in air is 

(1) ${\epsilon }_0$

(2) ${\epsilon }_0^{-1}$

(3) ${\left(4p{\epsilon }_0\right)}^{-1}$

(4) $4\pi {\epsilon }_0$ 

A cube of side l is placed in a uniform field E, where $E=E\hat{i}$. The net electric flux through the cube is

(1) Zero

(2) l²E

(3) 4l²E

(4) 6l²E

Eight dipoles of charges of magnitude \((e)\) are placed inside a cube. The total electric flux coming out of the cube will be: 
1. \(\frac{8e}{\epsilon _{0}}\)
2. \(\frac{16e}{\epsilon _{0}}\)
3. \(\frac{e}{\epsilon _{0}}\)
4. zero

A charge \(q\) is placed at the centre of the open end of the cylindrical vessel. The flux of the electric field through the surface of the vessel is:
1. \(0\)
2. \(\dfrac{q}{\varepsilon_0}\)
3. \(\dfrac{q}{2\varepsilon_0}\)
4. \(\dfrac{2q}{\varepsilon_0}\)

According to Gauss’ Theorem, electric field of an infinitely long straight wire is proportional to 

(1) r

(2) $\frac{1}{r^2}$

(3) $\frac{1}{r^3}$

(4) $\frac{1}{r}$

Electric charge is uniformly distributed along a long straight wire of radius \(1\) mm. The charge per cm length of the wire is \(Q\) coulomb. Another cylindrical surface of radius \(50\) cm and length \(1\) m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is:

The S.I. unit of electric flux is 

(1) Weber

(2) Newton per coulomb

(3) Volt × metre

(4) Joule per coulomb

\(q_1, q_2,q_3~\text{and}~q_4\) are point charges located at points as shown in the figure and \(S\) is a spherical Gaussian surface of radius \(R\). Which of the following is true according to the Gauss’s law?

1. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3\right) \cdot d \vec{A}=\frac{q_1+q_2+q_3}{2 \varepsilon_0}\)
2. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3+\vec{E}_4\right) \cdot d \vec{A}=\frac{\left(q_1+q_2+q_3\right)}{\varepsilon_0}\)
3. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3\right) \cdot d \vec{A}=\frac{\left(q_1+q_2+q_3+q_4\right)}{\varepsilon_0}\)
4. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3+\vec{E}_4\right) \cdot d \vec{A}=\frac{\left(q_1+q_2+q_3+q_4\right)}{\varepsilon_0}\)

If the electric flux entering and leaving an enclosed surface respectively is ${\phi }_1$ and ${\phi }_2$, the electric charge inside the surface will be: 

(1) $({\phi }_1+{\phi }_2){\epsilon }_0$

(2) $({\phi }_2-{\phi }_1){\epsilon }_0$

(3) $({\phi }_1+{\phi }_2)/{\epsilon }_0$

(4) $({\phi }_2-{\phi }_1)/{\epsilon }_0$