Question

calculate the amount of work done to dissociate a system of three charges 2 of 1 microcoulomb and one of - 4 microcoulomb place on the vertices of an equilateral triangle of side 10cm

Answer 1

Given:-

$q_{A}$ = 1×10⁻⁶C = $q_{B}$

$q_{C}$ = -4×10⁻⁶C

r = 10cm = 0.1m

To Find:-

The amount of work done to dissociate a system of three charges as shown in attached figure.

Solution:-

Therefore the total work done according to figure

W = $W_{AB} + W_{BC} + W_{AC}$

W = $\frac{1}{4\pi E_{0} r} [ q_{A}q_{B} + q_{B}q_{C} + q_{C}q_{A} ]$            Where   $\frac{1}{4\pi E_{0}}$ = 9x10⁹

W = $9x10^{9}(\frac{1}{0.1}) x [1x1-4x1-4x1 ]$

W = $\frac{-9x10^{9}(7x10^{-12}) }{0.1}$      By solving above equation, we get

W = -63Joule