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 K₁&K₂ ($K_1<K_2$) are inserted between plates of a parallel capacitor, as shown in the figure. The variation of electric field E between the plates with distance d as measured from plate P is correctly shown by  

1.

2. 

3.

4.

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

1. zero and Q/4π$\epsilon$_oR² 

2. Q/4π$\epsilon$_oR and zero

3. Q/4π$\epsilon$_oR and Q/4π$\epsilon$_oR²

4. Both are zero

Four point charges $-Q,-q,2q<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>2Q$ 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 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. $3C,3V$

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. \(\frac{2\varepsilon_0A}{d}\left(\frac{K_1\times K_2}{K_1+K_2}\right)\)