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 $V_P-V_Q \mathrm{and} V_B-V_A \mathrm{are} \mathrm{respectively}:$

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