Question

Two equal charges, 2.0 × 10−7 C each, are held fixed at a separation of 20 cm. A third charge of equal magnitude is placed midway between the two charges. It is now moved to a point 20 cm from both the charges. How much work is done by the electric field during the process?

Answer 1

Answer: -3.6 mili joule

Explanation:

We have to use the concept if electrostatic potential energy.

Answer 2

Work Done is $3.6 \times 10^-^3$ J

Explanation:

$E_1 = \frac {Kq_1q_2}{r} + \frac {Kq_3q_4}{r}$

= $\frac {9 \times 10^9(2 \times 10^-^7)^2}{0.1} + \frac {9 \times 10^9(2 \times 10^-^7)^2}{0.1}$

= $\frac {2 \times9 \times 10^9 \times 4 \times 10^{-14}}{0.1}$

= J

When charge is placed at B,

$E_2 = \frac {Kq_1q_2}{r} + \frac {Kq_3q_4}{r}$

= $\frac {2 \times9 \times 10^9 \times 4 \times 10^{-14}}{0.2}$

= J

Work Done  = $E_1 - E_2$

= $(72 - 36 \times 10^-^4)$

= J

= $3.6 \times 10^-^3$ J