Two charges are placed at a certain distance apart in the air. If a glass slab is introduced between them, then the force between the two charges will

1.  Become zero

2.  Decrease

3.  Increase

4.  Remain the same

A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is

1.  $qEy$

2.  $qE{y}^2$

3.  $q{E}^{2}y$

4.  $q^{2}Ey$

Force of interaction between two bodies of the same nature of the charge 

1.  Must be repulsive

2.  May be attractive

3.  Must be attractive

4.  Zero

On the axis of a uniformly charged ring of radius R, as shown, the electric field is maximum at

1.  x = 0

2.  $x = \frac{R}{2}$

3.  $x = R\sqrt{2}$

4.  $x = \frac{R}{\sqrt{2}}$

At the three corners of an equilateral triangle, charges are placed as shown in the figure. The magnitude of the electric field at point O is

1.  $\frac{1}{4\pi {\epsilon }_0}\frac{2Q}{r^2}$

2.  $\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$

3.  $\frac{1}{4\pi {\epsilon }_0}\frac{Q\sqrt{2}}{r^2}$

4.  $\frac{1}{4\pi {\epsilon }_0}\frac{Q\sqrt{3}}{r^2}$

Two symmetrical parallel metal each of one side plate area A having charges $Q_1<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>-Q_2$ are placed at small separation. What will be the electric field at a point in between the plates?

1.  $\frac{Q_1+Q_2}{2A{\epsilon }_0}$

2.  $\frac{Q_1+Q_2}{4A{\epsilon }_0}$

3.  $\frac{Q_1-Q_2}{2A{\epsilon }_0}$

4.  $\frac{Q_1-Q_2}{4A{\epsilon }_0}$

An electric dipole of the moment $\overrightarrow{P}$ is released in a uniform electric field $\overrightarrow{E}$ from the position of maximum torque. The angular speed of the dipole when $\overrightarrow{P}$ becomes parallel to $\overrightarrow{E}$ will be [l = moment of inertia of dipole]

1.  $\sqrt{\frac{2PE}{I}}$

2.  $\sqrt{\frac{PE}{I}}$

3.  $\sqrt{\frac{4PE}{I}}$

4.  $\sqrt{\frac{2I}{PE}}$

A charge Q is given to a conducting sphere of radius R. Now if a charge -Q is placed, as shown, at a distance r from the center, then the magnitude of the force of attraction between charges is

1.  $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{Q^2}{r^2}$

2.  <$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{Q^2}{r^2}$

3.  >$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{Q^2}{r^2}$

4.  $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{Q^2}{{\left(r-R\right)}^2}$

If the electric flux entering and leaving a closed surface are respectively of magnitude ${\varphi }_1<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\varphi }_2$, then the electric charge inside the surface will be

1.  ${\epsilon }_0\left({\varphi }_2-{\varphi }_1\right)$

2.  $\left({\varphi }_1-{\varphi }_2\right){\mathrm{/2\epsilon }}_0$

3.  ${\epsilon }_0\left({\varphi }_2+{\varphi }_1\right)$

4.  $\frac{{\varphi }_2-{\varphi }_1}{{\epsilon }_0}$

Electric field at an axial point of short dipole is  $\overrightarrow{E_1}$. If the electric field at the equatorial point of same dipole is $\overrightarrow{E_2}$, then which of the following is correct?

1.  $\overrightarrow{E_1}$ . $\overrightarrow{E_2}$ < 0

2.  $\overrightarrow{E_1}$ . $\overrightarrow{E_2}$ > 0

3.  $\overrightarrow{E_1}$ . $\overrightarrow{E_2}$ = 0

4.  $\overrightarrow{E_1}$ . $\overrightarrow{E_2}$ = $\overrightarrow{0}$