Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC equal to 2a. D and E are the mid points of BC and CA. The work done in taking a charge Q from D to E is 

(1)$\frac{eqQ}{8{\mathrm{\pi \epsilon }}_0\alpha }$                                                 

(2)$\frac{qQ}{4{\mathrm{\pi \epsilon }}_0\mathrm{\alpha }}$

(3)zero                                                     

(4)$\frac{3qQ}{4{\mathrm{\pi \epsilon }}_0\mathrm{\alpha }}$

Three concentric spherical shells have radii a, b and c (a<b<c) and have surface charge densities $\mathrm{\sigma }, -\mathrm{\sigma }$ and $\mathrm{\sigma }$ respectively. If ${\mathrm{V}}_{\mathrm{A}}, {\mathrm{V}}_{\mathrm{B}}$ and ${\mathrm{V}}_{\mathrm{C}}$ denote the potential of the three shells, if c=a+b, we have

(1) ${\mathrm{V}}_{\mathrm{C}}={\mathrm{V}}_{\mathrm{A}}\ne {\mathrm{V}}_{\mathrm{B}}$                                       

(2) ${\mathrm{V}}_{\mathrm{C}}={\mathrm{V}}_B\ne {\mathrm{V}}_{\mathrm{A}}$

(3) ${\mathrm{V}}_{\mathrm{C}}\ne {\mathrm{V}}_B\ne {\mathrm{V}}_A$                                       

(4) ${\mathrm{V}}_{\mathrm{C}}={\mathrm{V}}_B={\mathrm{V}}_A$

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 \(K_1\) 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) $\frac{C_2{V}_1-C_1{V}_2}{C_1+C_2}$