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Q.
Three charges each $20\mu C$ are placed at the corners of an equilateral triangle of side 0.4m The potential energy of the.system is
Electrostatic Potential and Capacitance
Solution:
$P . E , u =\frac{ kq _{1} q _{2}}{ r }$
Now total P.E. $= P \cdot E _{ AB }+ P \cdot E _{ BC }+ P \cdot E _{ AC }$
$=\frac{ k (20 \mu C )(20 \mu C )}{(0.4)} \times 3$
$=\frac{9 \times 10^{9} \times 20 \times 20 \times 10^{-12} \times 3 \times 10}{4}$
$u =27 \,J$
