A parallel plate capacitor of capacitance 20 $\mu$F is being charged by a voltage source whose potential is charging at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively:

1. zero, zero

2. zero, 60$\mu$A

3. 60$\mu$A, 60$\mu$A

4. 60$\mu$A, zero

A parallel plate condenser has a uniform electric field E (V/m) in the space between the plates. If the distance between the plates is d(m) and area of each plate is A(m²) , the energy (joule) stored in the condenser is

1.  $\frac{1}{2}\left({\mathrm{\epsilon }}_{\mathrm{o}}\right){\left(\mathrm{E}\right)}^2$

2.  $\frac{{\mathrm{E}}^2 \mathrm{Ad}}{{\mathrm{\epsilon }}_{\mathrm{o}}}$

3.  $\frac{1}{2}\left({\mathrm{\epsilon }}_{\mathrm{o}}\right) {\mathrm{E}}^2 \mathrm{Ad}$

4.  $\left({\mathrm{\epsilon }}_{\mathrm{o}}\right) \mathrm{EAd}$

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

1. decreases

2. does not change

3. becomes zero

4. increases

Two condensers, one of capacity C and the other of capacity C/2 are connected to a V volt battery, as shown. 
 
The work done in charging fully both the condensers is :

1. $2 C{V}^2$

2. $\frac{1}{4} C{V}^2$

3. $\frac{3}{4} C{V}^2$

4. $\frac{1}{2} C{V}^2$

An electric dipole of moment $\overrightarrow{p}$ is lying along a uniform electric field $\overrightarrow{E}$. The work done in rotating the dipole by 90 ° is :

1. $\sqrt{2}pE$

2. $\frac{2E}{2}$

3. 2pE

4. pE

The true statement is, on increasing the distance between the plates of a parallel plate condenser -

(1) The electric intensity between the plates will decrease

(2) The electric intensity between the plates will increase

(3) The electric intensity between the plates will remain unchanged

(4) The P.D. between the plates will decrease

In a certain charge distribution, all points having zero potential can be joined by a circle S. Points inside S have positive potential, and points outside S have a negative potential. A positive charge, which is free to move, is placed inside S .

(1) It will remain in equilibrium

(2) It can move inside S, but it cannot cross S

(3) It must cross S at some time

(4) It may move, but will ultimately return to its starting point

Two concentric, thin metallic spheres of radii ${\mathrm{R}}_1$ and ${\mathrm{R}}_2$ $\left({\mathrm{R}}_1 > {\mathrm{R}}_2\right)$ bear changes ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ respectively. Then the potential at distance r between ${\mathrm{R}}_1$ and ${\mathrm{R}}_2$ will be $\left(\mathrm{k}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)$

1.  $\mathrm{k} \left(\frac{{\mathrm{Q}}_1+{\mathrm{Q}}_2}{\mathrm{r}}\right)$                       

2.  $\mathrm{k} \left(\frac{{\mathrm{Q}}_1}{\mathrm{r}}+\frac{{\mathrm{Q}}_2}{{\mathrm{R}}_2}\right)$

3.  $\mathrm{k} \left(\frac{{\mathrm{Q}}_2}{\mathrm{r}}+\frac{{\mathrm{Q}}_1}{{\mathrm{R}}_1}\right)$                     

4.  $\mathrm{k} \left(\frac{{\mathrm{Q}}_1}{{\mathrm{R}}_1}+\frac{{\mathrm{Q}}_1}{{\mathrm{R}}_2}\right)$

A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If x is the separation between the plates, the time rate of change of electrostatic energy of the capacitor is proportional to:

(1) x^–2

(2) x

(3) x^–1

(4) x²

n identical condensers are joined in parallel and are charged to potential V. Now they are separated and joined in series. Then the total energy and potential difference of the combination will be 

(1) Energy and potential difference remain the same

(2) Energy remains the same and the potential difference is nV

(3) Energy increases n times and potential difference is nV

(4) Energy increases n times and potential difference remains the same