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

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. \(\frac{4}{7}\)

An electric field is given by ${\mathrm{E}}_{\mathrm{x}} = – 2{\mathrm{x}}^3 \mathrm{kN}/\mathrm{C}$. The potential of the point (1, –2), if potential of the point (2, 4) is taken as zero, is –

1.  $-7.5\times {10}^3 \mathrm{V}$                     

2.  $7.5\times {10}^3 \mathrm{V}$

3.  $-15\times {10}^3 \mathrm{V}$                     

4.  $15\times {10}^3 \mathrm{V}$

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