Two charges 8 mu C each are placed at the co | vedclass

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(A) The electric potential $V$ at a point due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$,where $k = 9 	imes 10^9 \ N \ m^

Vedclass · Accepted Answer

$7.2 	imes 10^5 \ V$

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(A) The electric potential $V$ at a point due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$,where $k = 9 	imes 10^9 \ N \ m^2 \ C^{-2}$.Since the triangle is equilateral with side $a = 0.2 \ m$,the distance from both corners $A$ and $B$ to corner $C$ is $r = 0.2 \ m$.The total potential at corner $C$ is the algebraic sum of the potentials due to charges at $A$ and $B$: $V_C = V_A + V_B$.$V_C = \frac{k q_A}{r} + \frac{k q_B}{r} = \frac{k}{r} (q_A + q_B)$.Substituting the values: $V_C = \frac{9 	imes 10^9}{0.2} (8 	imes 10^{-6} + 8 	imes 10^{-6}) \ V$.$V_C = \frac{9 	imes 10^9}{0.2} (16 	imes 10^{-6}) \ V$.$V_C = 9 	imes 10^9 	imes 80 	imes 10^{-6} \ V$.$V_C = 720 	imes 10^3 \ V = 7.2 	imes 10^5 \ V$.

Two charges $8 \ \mu C$ each are placed at the corners $A$ and $B$ of an equilateral triangle of side $0.2 \ m$ in air. The electric potential at the third corner $C$ is

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