Three charges of + 0.1 C each are placed at the corners of an equilateral triangle as shown in figure. If energy is supplied at the rate of 1 kW, how many days would be required to move the charge at A to a point D which is the mid-point of the line BC?

Solution

Given,
Charges q_A, q_B, q_C= 0.1 C
Sides of equilateral triangle, a = 1 m
Energy supplied, E = 1 kW

Potential at A due to charges at B and C is given by

V = \frac{1}{4 {πε}_{\circ}} \frac{q}{r} ∴ V_{A} = \frac{1}{4 {πε}_{0}} \frac{0.1}{1} + \frac{1}{4 {πε}_{0}} \frac{0.1}{1} = 2 \times 9 \times 10^{9} \times \frac{1}{10} volt = 18 \times 10^{8} volt

Potential at D due to charges at B and C is given by

V_{D} = \frac{1}{4 {πε}_{0}} \frac{0.1}{0.5} + \frac{1}{4 {πε}_{0}} \frac{0.1}{0.5}

= 2 \times 9 \times 10^{9} \times \frac{1}{5} V = 36 \times 10^{8} V

Now, potential difference between A and D is
V_D–V_A = (36 – 18) x 10⁸V
=1.8 x 10⁸V

∴

Work done in moving charge 0.1 C from A to D,
W = V.q
W = 0.1 C x 18 x 10⁸ V
= 1.8 x 10⁸ J

We know that

Power = \frac{Work}{Time} or Time = \frac{Work}{Power}

Time t taken to move the charge from A to D,

= \frac{1.8 \times 10^{8} J}{1 kW} = \frac{1.8 \times 10^{8} J}{10^{3} {Js}^{- 1}} = \frac{1.8 \times 10^{5}}{3600} h = 50 h = \frac{50}{24} = 2.08 d a y s

Therefore, 2 days will be required to move the charge from point A to point D.

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