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

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

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:
 

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

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:
 

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

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:

$1.<mspace\; width="1em"></mspace>\frac{-q^2}{4\pi {\epsilon }_{0}d}\left(4-\sqrt{2}\right)$
$2.<mspace\; width="1em"></mspace>\frac{-q^2}{4\pi {\epsilon }_{0}d}\left(2-\sqrt{2}\right)$
$3.<mspace\; width="1em"></mspace>\frac{-q^2}{4\pi {\epsilon }_{0}d}\left(4+\sqrt{2}\right)$
$4.<mspace\; width="1em"></mspace>\frac{-q^2}{4\pi {\epsilon }_{0}d}\left(2+\sqrt{2}\right)$

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. \(\frac{-q^2}{4\pi\varepsilon_0 d}(4-\sqrt{2})\)
2. zero
3. \(\frac{-q^2}{4\pi\varepsilon_0 d}(4+\sqrt{2})\)
4. \(\frac{-q^2}{4\pi\varepsilon_0 d}(2+\sqrt{2})\)

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