A parallel plate condenser is immersed in an oil of dielectric constant 2. The field between the plates is 

1. Increased proportional to 2

2. Decreased proportional to $\frac{1}{2}$

3. Increased proportional to $\sqrt{2}$

4. Decreased proportional to $\frac{1}{\sqrt{2}}$ 

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 \(x\).

When air in a capacitor is replaced by a medium of dielectric constant K, the capacity:
1. Decreases K times
2. Increases K times
3. Increases K² times
4. Remains constant

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 radii of two metallic spheres P and Q are r₁ and r₂ respectively. They are given the same charge. If r₁ > r₂  , then on connecting them with a thin wire, the charge will flow 

1. From P to Q

2. From Q to P

3. Neither the charge will flow from P to Q nor from Q to P

4. The information is incomplete

Between the plates of a parallel plate condenser, a plate of thickness t₁ and dielectric constant k₁ is placed. In the rest of the space, there is another plate of thickness t₂ and dielectric constant k₂. The potential difference across the condenser will be 

1. $\frac{Q}{A{\epsilon }_0}\left(\frac{{\displaystyle t_1}}{{\displaystyle k_1}}+\frac{{\displaystyle t_2}}{{\displaystyle k_2}}\right)$

2. $\frac{{\epsilon }_{0}Q}{A}\left(\frac{{\displaystyle t_1}}{{\displaystyle k_1}}+\frac{{\displaystyle t_2}}{{\displaystyle k_2}}\right)$

3. $\frac{Q}{A{\epsilon }_0}\left(\frac{{\displaystyle k_1}}{{\displaystyle t_1}}+\frac{{\displaystyle k_2}}{{\displaystyle t_2}}\right)$

4. $\frac{{\epsilon }_{0}Q}{A}(k_1{t}_1+k_2{t}_2)$

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

The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively C_o and W_o. If the air is replaced by the glass (dielectric constant = 5) between the plates, the capacity of the condenser and the energy stored in it will respectively be -

1. $5C_o,5W_o$

2. $5C_o,\frac{W_0}{5}$

3. $\frac{C_o}{5},5W_o$

4. $\frac{C_o}{5},\frac{W_o}{5}$

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

A 6 μF capacitor is charged from 10 volts to 20 volts. Increase in energy will be 

1. 18 × 10^–4 J

2. 9 × 10^–4 J

3. 4.5 × 10^–4 J

4. 9 × 10^–6 J