A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference V is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is

(1)C²V²/2d

(2)CV²/2d

(3)CV²/d

(4)C²V²/2d²

Two thin dielectric slabs of dielectric constants \(K_1~\text{and}~K_2(K_{1} < K_{2})\) are inserted between plates of a parallel capacitor, as shown in the figure. The variation of the electric field \(E\) between the plates with distance \(d\) as measured from the plate \(P\) is correctly shown by:  
   

1.		2.
3.		4.

Four point charges \(-Q, -q,2q~\text{and}~2Q\) are placed, one at each corner of the square. The relation between \(Q\) and \(q\) for which the potential at the center of the square is zero, is:

Two metallic spheres of radii 1 cm and 3 cm

are given charges of -1$\times {10}^{-2}\mathrm{C}$ and $5\times {10}^{-2}\mathrm{C}$,

respectively. If these are connected by a conducting

wire, the final charge on the bigger sphere is

(a) $2\times {10}^{-2}\mathrm{C}$

(b) $3\times {10}^{-2}\mathrm{C}$

(c) $4\times {10}^{-2}\mathrm{C}$

(d) $1\times {10}^{-2}\mathrm{C}$

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 $\mathrm{A}\left({\mathrm{m}}^2\right)$, the energy (joule) stored

in the condenser is

(a) $\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}^2$

(b) ${\mathrm{\epsilon }}_0\mathrm{EAd}$

(c) $\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}^2\mathrm{Ad}$

(d) ${\mathrm{E}}^2\mathrm{Ad}/{\mathrm{\epsilon }}_0$

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

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

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

(c)    Zero

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

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

1. $\frac{qQ}{8{\mathrm{\pi \epsilon }}_{0}a}$                                 
2. $\frac{qQ}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}$
3. zero                           
4. $\frac{3qQ}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}$

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. \(3C, 3V\)

Five identical plates each of area \(A\) are joined as shown in the figure. The distance between the plates is \(d\). The plates are connected to a potential difference of \(V\) volts. The charge on plates \(1\) and \(4\) will be:

        
1. \(-\frac{\varepsilon_{0} A V}{d} ,  \frac{2\varepsilon_{0} A V}{d}\)
2. \(\frac{\varepsilon_{0} A V}{d} ,  \frac{2\varepsilon_{0} A V}{d}\)
3. \(\frac{\varepsilon_{0} A V}{d} , -\frac{2\varepsilon_{0} A V}{d}\)
4. \(-\frac{\varepsilon_{0} A V}{d} ,  -\frac{2\varepsilon_{0} A V}{d}\)