The capacitance of a parallel plate capacitor is C. If a dielectric slab of thickness equal to one-fourth of the plate separation and dielectric constant K is inserted between the plates, then new capacitance become

1. $\frac{KC}{2\left(K+1\right)}$

2. $\frac{2KC}{K+1}$

3. $\frac{5KC}{4K+1}$

4. $\frac{4KC}{3K+1}$

The electric potential at a point at distance 'r' from a short dipole is proportional to

(1) $r^2$

(2) ${\mathrm{r}}^{-1}$

(3) $r^{-2}$

(4) $r^1$

A hollow charged metal spherical shell has radius R. If the potential difference between its surface and a point at a distance 3R from the center is V, then the value of electric field intensity at a point at distance 4R from the center is

1.  $\frac{3\mathrm{V}}{19\mathrm{R}}$

2.  $\frac{\mathrm{V}}{6\mathrm{R}}$

3.  $\frac{3\mathrm{V}}{32\mathrm{R}}$

4.  $\frac{3\mathrm{V}}{16\mathrm{R}}$

You are traveling in a car during a thunderstorm. In order to protect yourself from lightning, you would prefer to

(1) Touch the nearest electric pole

(2) Take shelter under a tree

(3) Get out and crouch on the ground

(4) Remain in the car

Capacitors $C_1=10\mu F and C_2=30\mu F$ are connected in series across a source of emf 20KV. The potential difference across $C_1$ will be

(1) 5 KV

(2) 15 KV

(3) 10 KV

(4) 20 KV

The equivalent capacitance between A and B is as the given figure:

(1) $16\mathrm{\pi }{\in }_0\mathrm{r}$

(2) $4\mathrm{\pi }{\in }_0\mathrm{r}$

(3) $8\mathrm{\pi }{\in }_0\mathrm{r}$

(4) None of these

Two metallic spheres of radii 2cm and 3cm are given charges 6mC and 4mC respectively. The final charge on the smaller sphere will be if they are connected by a conducting wire

(1) 4mC

(2) 6mC

(3) 5mC

(4) 10mC

When a proton at rest is accelerated by a potential difference \(V\), its speed is found to be \(v\). The speed of an \(\alpha-\textit{particle}\) when accelerated by the same potential difference from rest will be: 

In the circuit shown in figure, energy stored in 6 $\mathrm{\mu F}$ capacitor will be:
 

The figure shows some of the equipotential surfaces. Magnitude and direction of the electric field is given by

1.  200 V/m, making an angle ${120}^0$ with the x-axis

2. 100 V/m, pointing towards the negative x-axis

3. 200 V/m, making an angle $-{60}^0$ with the x-axis

4. 100 V/m, making an angle ${30}^0$ with the x-axis