The potential at a point P due to a charge of 4 × 10^–7 C located 9 cm away is:

$1. 3\times {10}^6 V$
$2. 4\times {10}^6 V$
$3. 3\times {10}^4 V$
$4. 4\times {10}^4 V$

The potential at point P is $4\times {10}^4 V$. The work done in bringing a charge of 2 × 10^–9 C from infinity to the point P is:

$1. 6\times {10}^{-6} J$
$2. 4\times {10}^{-4} J$
$3. 8\times {10}^{-5} J$
$4. 9\times {10}^{-4} J$

Two charges 3 × 10^–8 C and –2 × 10^–8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

1. 9 cm away from 3 × 10^–8 C

2. 25 cm away from 3 × 10^–8 C

3. 45 cm away from 3 × 10^–8 C

4. Both (1) and (3)

Figures (a) and (b) show the field lines of a positive and negative point charge respectively.

The signs of the potential difference $V_P-V_Q \mathrm{and} V_B-V_A \mathrm{are} \mathrm{respectively}:$

$1. +, -$
$2. +, +$
$3. -, +$
$4. -, -$

Figures (a) and (b) show the field lines of a positive and negative point charge respectively. The signs of the potential energy difference of a small negative charge between the points Q and P & A and B are respectively:
 

$1. +, -$
$2. +, +$
$3. -, +$
$4. -, -$

Figures (a) and (b) show the field lines of a positive and negative point charge respectively. The sign of the work done by the field in moving a small positive charge from Q to P and the sign of the work done by the external agency in moving a small negative charge from B to A, respectively, will be:
 

$1. +, -$
$2. +, +$
$3. -, +$
$4. -, -$

Figures (a) and (b) show the field lines of a positive and negative point charge respectively. The kinetic energy of a small negative charge in going from B to A:

1. decreases.

2. increases.

3. remains the same.

4. first increases and then decreases.

Four charges are arranged at the corners of a square ABCD of side d, as shown in the figure. The work required to put together this arrangement will be:

$1. \frac{-q^2}{4\pi {\epsilon }_{0}d}\left(4-\sqrt{2}\right)$
$2. \frac{-q^2}{4\pi {\epsilon }_{0}d}\left(2-\sqrt{2}\right)$
$3. \frac{-q^2}{4\pi {\epsilon }_{0}d}\left(4+\sqrt{2}\right)$
$4. \frac{-q^2}{4\pi {\epsilon }_{0}d}\left(2+\sqrt{2}\right)$

Four charges are arranged at the corners of a square ABCD of side d, as shown in the figure. A charge $q_0$ is brought from $\infty$ to the centre E of the square, the four charges being held fixed at its corners. How much work is needed to do this?

$1. \frac{-{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{d}}\left(4-\sqrt{2}\right)$
$2. \mathrm{zero}$
$3. \frac{-{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{d}}\left(4+\sqrt{2}\right)$
$4. \frac{-{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{d}}\left(2+\sqrt{2}\right)$

The electrostatic potential energy of a system consisting of two charges 7 µC and –2 µC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively is:

1. 0.2 J

2. -0.7 J

3. -0.2 J

4. 0.7 J