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(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}$.Since the triang

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$36 	imes {10^4}\,V$

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(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}$.Since the triangle is equilateral with side length $0.2\, m$,the distance from both charges at $A$ and $B$ to point $C$ is $r = 0.2\, m$.The total potential at $C$ is the sum of potentials due to both charges:$V_C = V_A + V_B = \frac{1}{4\pi \varepsilon_0} \frac{q_A}{r} + \frac{1}{4\pi \varepsilon_0} \frac{q_B}{r}$Given $q_A = q_B = 4\,\mu C = 4 	imes 10^{-6}\, C$ and $r = 0.2\, m$:$V_C = 2 	imes \left( 9 	imes 10^9 	imes \frac{4 	imes 10^{-6}}{0.2} ight)$$V_C = 2 	imes (9 	imes 10^9 	imes 20 	imes 10^{-6})$$V_C = 2 	imes 18 	imes 10^4 = 36 	imes 10^4\, V$.

Two charges of $4\,\mu C$ each are placed at the corners $A$ and $B$ of an equilateral triangle of side length $0.2\, m$ in air. Find the electric potential at $C$. $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\text{N m}^2/\text{C}^2} \right]$

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