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 \(K_1\) and \(K_2\) have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor? 

1. \(\frac{2\varepsilon_0A}{2}\left(K_1+K_2\right)\)
2. \(\frac{2\varepsilon_0A}{2}\frac{\left(K_1+K_2\right)}{K_1\times K_2}\)
3. \(\frac{2\varepsilon_0A}{d}\left(\frac{K_1+K_2}{K_1-K_2}\right)\)
4. \(\frac{2\varepsilon_0A}{d}\left(\frac{K_1\times K_2}{K_1+K_2}\right)\)

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

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

In the given network capacitance, C₁ = 10 μF, C₂ = 5 μF and C₃ = 4 μF. What is the resultant capacitance between A and B 

(1) 2.2 μF

(2) 3.2 μF

(3) 1.2 μF

(4) 4.7 μF

The equivalent capacitance between \(A\) and \(B\) is: