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

Shown below is a distribution of charges. The flux of electric field due to these charges through the surface S is 

(1) $3q/{\epsilon }_0$

(2) $2q/{\epsilon }_0$

(3) $q/{\epsilon }_0$ 

(4) Zero

Consider the charge configuration and spherical Gaussian surface as shown in the figure. While calculating the flux of the electric field over the spherical surface, the electric field will be due to: 

(1) q₂ only

(2) Only the positive charges

(3) All the charges

(4) +q₁ and – q₁ only