A conducting sphere of radius R is given a charge Q. The electric potential and field at the centre of the sphere respectively are

(a) zero and Q/4π$\epsilon$_oR² 

(b)Q/4π$\epsilon$_oR and zero

(c)Q/4π$\epsilon$_oR and Q/4π$\epsilon$_oR²

(d)Both are zero

In a region, the potential is represented by V(x,y,z)=6x-8xy-8y+6yz, where V is in volts and x,y,z are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1,1,1) is

(1)6√5N

(2)30N

(3)24N

(4)4√35N

\(A,B\) and \(C\) are three points in a uniform electric field. The electric potential is:
               

Four point charges $-Q,-q,2q and 2Q$ are placed, one at each corner of the square.The relation between Q and q for which the potential at the centre of the square is zero, is 

(1) Q=-q                                       

(2)Q=-$\frac{1}{q}$

(3)Q=q                                         

(4)Q=$\frac{1}{q}$

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

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) $3C,3V$