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

A series combination of \(n_1\) capacitors, each of value \(C_1\), is charged by a source of potential difference \(4\) V. When another parallel combination of \(n_2\) capacitors, each of value \(C_2\), is charged by a source of potential difference \(V\), it has the same (total) energy stored in it as the first combination has. The value of \(C_2\) in terms of \(C_1\) is:
1. \(\frac{2C_1}{n_1n_2}\)
2. \(16\frac{n_2}{n_1}C_1\)
3. \(2\frac{n_2}{n_1}C_1\)
4. \(\frac{16C_1}{n_1n_2}\)

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