Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is 

1. 8 times

2. 4 times

3. 2 times

4. 32 times

The capacity of a parallel plate condenser is C. It's capacity when the separation between the plates is halved will be?

1. 4 C

2. 2 C

3. $\frac{C}{2}$

4. $\frac{C}{4}$

If the dielectric constant and dielectric strength be denoted by \(k\) and \(x\) respectively, then a material suitable for use as a dielectric in a capacitor must have:
1. high \(k\) and high \(x\).
2. high \(k\) and low \(x\).
3. low \(k\) and low \(x\).
4. low \(k\) and high \(x\).

64 drops each having the capacity C and potential V are combined to form a big drop. If the charge on the small drop is q, then the charge on the big drop will be 

1. 2q

2. 4q

3. 16q

4. 64q

The capacity of a parallel plate capacitor increases with the 

1. Decrease of its area

2. Increase of its distance

3. Increase of its area

4. None of the above

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

Force of attraction between the plates of a parallel plate capacitor whose dielectric constant is K will be -

1. $\frac{q^2}{2{\epsilon }_{0}AK}$

2. $\frac{q^2}{{\epsilon }_{0}AK}$

3. $\frac{q}{2{\epsilon }_{0}A}$

4. $\frac{q^2}{2{\epsilon }_0{A}^{2}K}$

The intensity of electric field at a point between the plates of a charged capacitor 

1. Is directly proportional to the distance between the plates

2. Is inversely proportional to the distance between the plates

3. Is inversely proportional to the square of the distance between the plates

4. Does not depend upon the distance between the plates

A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes 

1. $\sqrt{2}C$

2. 2C

3. $\frac{C}{\sqrt{2}}$

4. $\frac{C}{2}$