A dipole is placed in an electric field as shown. In which direction will it move? 

   

1.	towards the left as its potential energy will decrease.
2.	towards the right as its potential energy will increase.
3.	towards the left as its potential energy will increase.
4.	towards the right as its potential energy will decrease.

Polar molecules are the molecules:

A charge q is located at the centre of a cube. The electric flux through any face is:

1. $\frac{2\mathrm{\pi q}}{6\left(4{\mathrm{\pi \epsilon }}_0\right)}$

2. $\frac{4\mathrm{\pi q}}{6\left(4{\mathrm{\pi \epsilon }}_0\right)}$

3. $\frac{\mathrm{\pi q}}{6\left(4{\mathrm{\pi \epsilon }}_0\right)}$

4. $\frac{\mathrm{q}}{6\left(4{\mathrm{\pi \epsilon }}_0\right)}$

A dipole of moment $\overrightarrow{p}$ is placed in uniform electric field $\overrightarrow{E}$. The torque acting on it is given by:
1. \(\vec{\tau }=\vec{p}.\vec{E}\)
2. \(\vec{\tau }=\vec{p} \times\vec{E}\)
3. \(\vec{\tau }=\vec{p}+\vec{E}\)
4. \(\vec{\tau }=\vec{p}-\vec{E}\)

If a charge Q µC is placed at the centre of the cube, the flux (In SI unit) coming out from any surface will be: 

1.  $\frac{Q}{6{\epsilon }_0}$ $\times$ ${10}^{-6}$

2.  $\frac{Q}{6{\epsilon }_0}$ $\times$ ${10}^{-3}$

3.  $\frac{Q}{2{\epsilon }_0}$ $$

4. $\frac{Q}{8{\epsilon }_0}$ $$

In Millikan oil drop experiment, a charged drop falls with a terminal velocity v. If an electric field E is applied vertically upwards it moves with terminal velocity 2v in upward direction. If electric field reduces to E/2 then its terminal velocity will be:
1. v/2
2. v
3. 3v/2
4. 2v

The unit of permittivity of free space ε₀ is:

A point \(Q\) lies on the perpendicular bisector of an electric dipole of dipole moment \(p.\) If the distance of \(Q\) from the dipole is \(r\) (much larger than the size of the dipole), then the electric field at \(Q\) is proportional to:
1. \(p^{2}\) and \(r^{-3}\)
2. \(p\) and \(r^{-2}\)
3. \(p^{-1}\) and \(r^{-2}\)
4. \(p\) and \(r^{-3}\)

The electric field at the equator of a dipole is \(E.\) If the strength of the dipole and distance are now doubled, then the electric field will be:

A charge \(q\) is placed in a uniform electric field \(E.\) If it is released, then the kinetic energy of the charge after travelling distance \(y\) will be:
1. \(qEy\)
2. \(2qEy\)
3. $\frac{qEy}{2}$
4. $\sqrt{qEy}$