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.

The electric field vector in a region given by $\overrightarrow{E}=(3\hat{i}+4y\hat{j})V{m}^{-1}$. Calculate the potential at (1m, 1m) taking potential at origin to be zero. [This question includes concepts from 12th syllabus]

1. 5V

2. 3V

3. -1V

4. -5V

Two-point charges each of charge +q are fixed at  (+a, 0) and (-a, 0).  Another positive point charge q placed at the origin is free to move along X-axis. The charge q at origin in equilibrium will have

1. maximum force and minimum potential energy.

2. minimum force and maximum potential energy.

3. maximum force and maximum potential energy.

4. minimum force and minimum potential energy.

An electric dipole of moment \(\vec {p} \) is lying along a uniform electric field \(\vec{E}\). The work done in rotating the dipole by \(90^{\circ}\) is:
1. \(\sqrt{2}pE\)
2. \(\frac{pE}{2}\)
3. \(2pE\)
4. \(pE\)

Charges +q and –q are placed at points A and B, respectively; which are at a distance 2L apart. C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is:
   
1. $\frac{qQ}{4{\mathrm{\pi \epsilon }}_0\mathrm{L}}$
2. $\frac{qQ}{2{\mathrm{\pi \epsilon }}_0\mathrm{L}}$
3. $\frac{qQ}{6{\mathrm{\pi \epsilon }}_0\mathrm{L}}$
4. $-\frac{qQ}{6{\mathrm{\pi \epsilon }}_0\mathrm{L}}$

Two condensers, one of capacity \(C\) and the other of capacity \(\frac{C}2\) are connected to a \(V\) volt battery, as shown in the figure. 
          
The energy stored in the capacitors when both condensers are fully charged will be:
1. \(2CV^2\)
2. \({1 \over4}CV^2\)
3. \({3 \over4}CV^2\)
4. \({1 \over2}CV^2\)

The electric potential at a point in free space due to a charge \(Q\) coulomb is \(Q\times10^{11}~\text{V}\). The electric field at that point is:
1. \(4\pi \varepsilon_0 Q\times 10^{22}~\text{V/m}\)
2. \(12\pi \varepsilon_0 Q\times 10^{20}~\text{V/m}\)
3. \(4\pi \varepsilon_0 Q\times 10^{20}~\text{V/m}\)
4. \(12\pi \varepsilon_0 Q\times 10^{22}~\text{V/m}\)

Three capacitors each of capacitance \(C\) and of breakdown voltage \(V\) are joined in series. The capacitance and breakdown voltage of the combination will be:
1. $\frac{C}{3},$ $\frac{V}{3}$

2. $3C,$ $\frac{V}{3}$

3. $\frac{C}{3},$ $3V$

4. \(3C,~3V\)

An electric dipole of moment \(p\) is placed in an electric field of intensity \(E.\) The dipole acquires a position such that the axis of the dipole makes an angle \(\theta\) with the direction of the field. Assuming that the potential energy of the dipole to be zero when \(\theta = 90^{\circ}\)$$, the torque and the potential energy of the dipole will respectively be:
1. \(pE\text{sin}\theta, ~-pE\text{cos}\theta\)
2. \(pE\text{sin}\theta, ~-2pE\text{cos}\theta\)
3. \(pE\text{sin}\theta, ~2pE\text{cos}\theta\)
4. \(pE\text{cos}\theta, ~-pE\text{sin}\theta\)