A charge of 40 μC is given to a capacitor having capacitance C = 10 μF. The stored energy in ergs is 

1. 80 × 10^–6

2. 800

3. 80

4. 8000

A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference V₀. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is 

1. $\frac{3{\epsilon }_{0}A{V}_0^2}{d}$

2. $\frac{{\epsilon }_{0}A{V}_0^2}{2d}$

3. $\frac{{\epsilon }_{0}A{V}_0^2}{3d}$

4. $\frac{{\epsilon }_{0}A{V}_0^2}{d}$

The electric field between the plates of a parallel plate capacitor when connected to a certain battery is E₀. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant K without disturbing the battery connections, the field between the plates shall be 

1. K E₀

2. E₀

3. $\frac{E_0}{K}$

4. None of the above

If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance 

1. Decreases two times

2. Increases two times

3. Increases four times

4. Remain the same

If there are n capacitors each of capacitance C in parallel connected to V volt source, then the energy stored is equal to 

1. CV

2. $\frac{1}{2}nC{V}^2$

3. CV²

4. $\frac{1}{2n}C{V}^2$

The unit of electric permittivity is 

1. Volt/m²

2. Joule/coulomb

3. Farad/m

4. Henry/m

The work done in placing a charge of 8 × 10^–18 coulomb on a condenser of capacity 100 micro-farad is 

1. 32× 10^–32 Joule

2. 16 × 10^–32 Joule

3. 3.1 × 10^–26 Joule

4. 4 × 10^–10 Joule

Two identical capacitors are joined in parallel, charged to a potential V and then separated and then connected in series i.e. the positive plate of one is connected to negative of the other 

1. The charges on the free plates are destroyed

2. The charges on the free plates are enhanced

3. The energy stored in the system increases

4. The potential difference in the free plates becomes 2V

A parallel plate capacitor is made by stacking n equally spaced plates connected alternately. If the capacitance between any two plates is C then the resultant capacitance is

1. C

2. nC

3. (n – 1)C

4. (n + 1)C