Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be

(1) $\frac{C}{3},\frac{V}{3}$

(2) $3C,\frac{V}{3}$

(3) $\frac{C}{3},3V$

(4) $3C,3V$

The electric potential at a point (x,y,z) is given by 

            $V=-x^{2}y-x{z}^3+4$

The electric field $\overrightarrow{E}$ at that point is 

(a) $\overrightarrow{E}=\hat{i}\left(2 xy+z^3\right)+\hat{j}{x}^2+\hat{k}3x{z}^2$

(b) $\overrightarrow{E}=\hat{i}2xy+\hat{j}\left(x^2+y^2\right)+\hat{k}\left(3xz-y^2\right)$

(c) $\overrightarrow{E}=\hat{i}{z}^3+\hat{j}xyz+\hat{k}{z}^2$

(d) $\overrightarrow{E}=\hat{i}\left(2xy-z^3\right)+\hat{j}x{y}^2+\hat{k}3z^{2}x$

The mean free path of electrons in a metal is $4\times {10}^{-8}m.$The electric field which can give on an average 2 eV energy to an electron in the metal will be in a unit of $V{m}^{-1}$ :

(1) $8\times {10}^7$                             

(2) $5\times {10}^{-11}$

(3) $8\times {10}^{-11}$                         

(4) $5\times {10}^7$

Two dielectric slabs of constant K₁ and K₂ have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor 

(1) $\frac{2{\epsilon }_{0}A}{2}(K_1+K_2)$

(2) $\frac{2{\epsilon }_{0}A}{2}\left(\frac{{\displaystyle K_1+K_2}}{{\displaystyle K_1\times K_2}}\right)$

(3) \(\frac{2\epsilon _{0}A}{d}(\frac{k_{1}+k_{2}}{k_{1}-k_{2}})\)

(4) $\frac{2{\epsilon }_{0}A}{d}\left(\frac{{\displaystyle K_1\times K_2}}{{\displaystyle K_1+K_2}}\right)$

What is the equivalent capacitance between A and B in the given figure (all are in farad) 

(1) $\frac{13}{18}F$

(2) $\frac{48}{13}F$

(3) $\frac{1}{31}F$

(4) $\frac{240}{71}F$

100 capacitors each having a capacity of 10 μF are connected in parallel and are charged by a potential difference of 100 kV. The energy stored in the capacitors and the cost of charging them, if electrical energy costs 108 paise per kWh, will be?

Four capacitors are connected as shown in the figure. Their capacities are indicated in the figure. The effective capacitance between points x and y is (in μF) 

(1) $\frac{5}{6}$

(2) $\frac{7}{6}$

(3) $\frac{8}{3}$

(4) 2

A 10 μF capacitor and a 20 μF capacitor are connected in series across a 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor 

(1) $\frac{400}{9}V$

(2) $\frac{800}{9}V$

(3) 400 V

(4) 200 V

Two condensers C₁ and C₂ in a circuit are joined as shown in figure. The potential of point A is V₁ and that of B is V₂. The potential of point D will be 

(1) $\frac{1}{2}(V_1+V_2)$

(2) $\frac{C_2{V}_1+C_1{V}_2}{C_1+C_2}$

(3) $\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

(4) $\frac{C_2{V}_1-C_1{V}_2}{C_1+C_2}$

The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that 

(1) Their capacities are equal

(2) Their capacities are 1 μF and 2 μF

(3) Their capacities are 0.5 μF and 1 μF

(4) Their capacities are infinite