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

$1.<mspace\; width="1em"></mspace>3\times {10}^6<mspace\; width="1em"></mspace>V$
$2.<mspace\; width="1em"></mspace>4\times {10}^6<mspace\; width="1em"></mspace>V$
$3.<mspace\; width="1em"></mspace>3\times {10}^4<mspace\; width="1em"></mspace>V$
$4.<mspace\; width="1em"></mspace>4\times {10}^4<mspace\; width="1em"></mspace>V$

The potential at point P is $4\times {10}^4<mspace\; width="1em"></mspace>V$. The work done in bringing a charge of 2 × 10^–9 C from infinity to the point P is:

$1.<mspace\; width="1em"></mspace>6\times {10}^{-6}<mspace\; width="1em"></mspace>J$
$2.<mspace\; width="1em"></mspace>4\times {10}^{-4}<mspace\; width="1em"></mspace>J$
$3.<mspace\; width="1em"></mspace>8\times {10}^{-5}<mspace\; width="1em"></mspace>J$
$4.<mspace\; width="1em"></mspace>9\times {10}^{-4}<mspace\; width="1em"></mspace>J$

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 $V_P-V_Q<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>V_B-V_A<mspace\; width="1em"></mspace>\mathrm{are}<mspace\; width="1em"></mspace>\mathrm{respectively}:$

$1.<mspace\; width="1em"></mspace>+,<mspace\; width="1em"></mspace>-$
$2.<mspace\; width="1em"></mspace>+,<mspace\; width="1em"></mspace>+$
$3.<mspace\; width="1em"></mspace>-,<mspace\; width="1em"></mspace>+$
$4.<mspace\; width="1em"></mspace>-,<mspace\; width="1em"></mspace>-$