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. $3\mathrm{C},<mspace\; width="1em"></mspace>\frac{\mathrm{V}}{3}$

2. $\frac{\mathrm{C}}{3}$, 3V

3. 3C, 3V

4. $\frac{C}{3},<mspace\; width="1em"></mspace>\frac{V}{3}$

A parallel plate condenser has a uniform electric field E ($\mathrm{V}<mspace\; width="1em"></mspace>{\mathrm{m}}^{-1}$) in the space between the plates. If the distance between the plates is d (m) and area of each plate is A $\left({\mathrm{m}}^2\right)$, the energy (joules) stored in the condenser is

1. $\frac{1}{2}{\epsilon }_0{E}^{2}Ad$

2. $E^{2}Ad/{\epsilon }_0$

3. $\frac{1}{2}{\epsilon }_0{E}^2$

4. ${\mathrm{\epsilon }}_0\mathrm{EAd}$

Three charges each +q are placed at the corners of an isosceles triangle ABC of sides BC and AC equal to 2a. D and E are the midpoints of BC and CA. The work done in taking a charge Q from D to E is

1. $\frac{3qQ}{4\pi {\epsilon }_{0}a}$

2. $\frac{3qQ}{8\pi {\epsilon }_{0}a}$

3. $\frac{qQ}{4\pi {\epsilon }_{0}a}$

4. zero

A, B and C are three points in a uniform electric field. The electric potential is

1. maximum at A

2. maximum at B

3. maximum at C

4. same at all the three points A, B and C

A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are

1. zero and $\frac{Q}{4\pi {\epsilon }_0{R}^2}$

2. $\frac{Q}{4\pi {\epsilon }_{0}R}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>zero$

3. $\frac{Q}{4\pi {\epsilon }_{0}R}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>\frac{Q}{4\pi {\epsilon }_0{R}^2}$

4. Both are zero

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

1. The energy stored in the capacitor decreases K times.

2. The change in energy stored is $\frac{1}{2}C{V}^2\left(\frac{1}{K}-1\right)$

3. The charge on the capacitor is not conserved.

4. The potential difference between the plates decreases K times.

A parallel plate air capacitor has capacity 'C', the distance of separation between plates is 'd' and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is :

1. $\frac{C^2{V}^2}{2d^2}$

2. $\frac{C^2{V}^2}{2d}$

3. $\frac{C{V}^2}{2d}$

4. $\frac{C{V}^2}{d}$

Two condensers one of capacity C and 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. $\frac{1}{4}C{V}^2$

2. $\frac{3}{2}C{V}^2$

3. $\frac{1}{2}C{V}^2$

4. $2{\mathrm{CV}}^2$

Charges +q and -q are placed at points  A and B respectively which are 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. $\frac{qQ}{2\pi {\epsilon }_{0}L}$

2. $\frac{qQ}{\mathit{6}\pi {\epsilon }_{0}L}$

3. $-\frac{qQ}{\mathit{6}\pi {\epsilon }_{0}L}$

4. $\frac{qQ}{\mathit{4}\pi {\epsilon }_{0}L}$

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.