A parallel plate condenser has a capacitance \(50~\mu\text{F}\) in air and \(110~\mu\text{F}\) when immersed in an oil. The dielectric constant \(k\) of the oil is: 
1. \(0.45\)
2. \(0.55\)
3. \(1.10\)
4. \(2.20\)

Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and thickness t(t < d) is introduced between the plates, its capacitance becomes -

(1) $\frac{{\epsilon }_{0}A}{d+t\left(1-\frac{1}{k}\right)}$

(2) $\frac{{\epsilon }_{0}A}{d+t\left(1+\frac{1}{k}\right)}$

(3) $\frac{{\epsilon }_{0}A}{d-t\left(1-\frac{1}{k}\right)}$

(4) $\frac{{\epsilon }_{0}A}{d-t\left(1+\frac{1}{k}\right)}$

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 glass (dielectric constant = 5) between the plates, the capacity of the plates and the energy stored in it will respectively be -

(1) $5C_o,5W_o$

(2) $C_o,\frac{W_0}{5}$

(3) $5C_o,\frac{W_o}{5}$

(4) $\frac{C_o}{5},\frac{W_o}{5}$

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

Three capacitors are connected to D.C. source of 100 volts shown in the adjoining figure. If the charge accumulated on plates of C₁, C₂ and C₃ are $q_a,q_b,q_c,q_d.q_e$ and q_f respectively, then 

(1) $q_b+q_d+q_f=\frac{100}{9}C$

(2) $q_b+q_d+q_f=0$

(3) $q_a+q_c+q_e=50C$

(4) $q_b=q_d=q_f$

Five capacitors of 10 μF capacity each are connected to a d.c. potential of 100 volts as shown in the adjoining figure. The equivalent capacitance between the points A and B will be equal to 

(1) 40 μF

(2) 20 μF

(3) 30 μF

(4) 10 μF

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system 

1. increases by a factor of 4

2.decreases by a factor of 2

3. remain the same 

4. increases by a factor of 2

The diagrams below show regions of equipotentials.

A positive charge is moved from A to B in each diagram. 

(1) Maximum work is requried to move q in figure(iii)

(2) In all the four cases,the work done is the same 

(3) Minimum work is requried to move q in the figure(i) 

(4) Maximum work is required to move q in figure(ii) 

A parallel-plate capacitor of area A, plate separation d, and capacitance C is filled with four dielectric materials having dielectric constants $k_1,k_2,k_{3 }$and $k_4$ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

(a) $k=k_1+k_2+k_3+3k_4$

(b) $k=\frac{2}{3}\left(k_1+k_2+k_3\right)+2k_4$

(c) $\frac{2}{k}=\frac{3}{k_1+k_2+k_3}+\frac{1}{k_4}$

(d) $\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+\frac{3}{2k_4}$

A capacitor of 2$\mu F$ is charged as shown in the figure. When the switch S is turned to position 2, the percentage of its stored energy dissipated is

1. 20%               2. 75%

3. 80%               4. 0%