The grid (each square of 1m × 1m), represents a region in space containing a uniform electric field.

If potentials at points O, A, B, C, D, E, F and G, H are respectively 0, –1, –2, 1, 2, 0, –1, 1 and 0 volts, find the electric field intensity –

1. $\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right) \mathrm{V}/\mathrm{m}$                           

2. $\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}\right) \mathrm{V}/\mathrm{m}$

3. $\left(-\hat{\mathrm{i}}+\hat{\mathrm{j}}\right) \mathrm{V}/\mathrm{m}$                         

4. $\left(-\hat{\mathrm{i}}-\hat{\mathrm{j}}\right) \mathrm{V}/\mathrm{m}$

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. $6\sqrt{5} N$

2. 30 N

3. 24 N

4. $4\sqrt{35} N$

An electric dipole of moment $p$ is placed in an electric field of intensity $E$. The dipole acquires a position such that the axis of the dipole makes an angle $\theta$ with the direction of the field. Assuming that the potential energy of the dipole to be zero when $\theta = 90^{\circ},$ the torque and the potential energy of the dipole will respectively be:

A network of four capacitors of capacity equal to $C_1 = C, C_2 = 2C, C_3 = 3C$ and $C_4 = 4C$ are connected in a battery as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is:
         
1. $\frac{22}{3}$
2. $\frac{3}{22}$
3. $\frac{7}{4}$
4. $\frac{4}{7}$

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. $-\frac{\varepsilon_{0} A V}{d} ,  -\frac{2\varepsilon_{0} A V}{d}$

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. $3C, 3V$

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. $\frac{3qQ}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}$

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

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