Two metallic spheres of radii $1$ cm and $3$ cm are given charges of $-1\times 10^{-2}~\text{C}$ and $5\times 10^{-2}~\text{C},$ respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is:
1. $2\times 10^{-2}~\text{C}$
2. $3\times 10^{-2}~\text{C}$
3. $4\times 10^{-2}~\text{C}$
4. $1\times 10^{-2}~\text{C}$

A parallel plate condenser has a uniform electric field $E$(V/m) in the space between the plates. If the distance between the plates is $d$(m) and area of each plate is $A(\text{m}^2)$, the energy (joule) stored in the condenser is:

Four electric charges +q, + q, -q and -q are placed at the corners of a square of side 2L (see figure). The electric potential at point A, mid-way between the two charges +q and +q, is

(1)  $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2\mathrm{q}}{\mathrm{L}}\left(1+\frac{1}{\sqrt{5}}\right)$

(2)    $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2\mathrm{q}}{\mathrm{L}}\left(1-\frac{1}{\sqrt{5}}\right)$

(3)    Zero

(4)  $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2\mathrm{q}}{\mathrm{L}}\left(1+\sqrt{5}\right)$

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. $5\times {10}^7$