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%

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 potential difference between the plates decreases K times

(2) The energy stored in the capacitor decreases K times

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

(4) The charge on the capacitor is not conserved

A parallel plate air capacitor has capacity C, 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)C²V²/2d

(2)CV²/2d

(3)CV²/d

(4)C²V²/2d²

Two thin dielectric slabs of dielectric constants \(K_1~\text{and}~K_2(K_{1} < K_{2})\) are inserted between plates of a parallel capacitor, as shown in the figure. The variation of electric field \(E\) between the plates with distance \(d\) as measured from plate \(P\) is correctly shown by:  
   

1.		2.
3.		4.