The potential difference (V_A-V_B) in the arr | vedclass

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Question

(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$, where $k = 9 	imes 10^9 	ext{ N m}^2/	ext{

Vedclass · Accepted Answer

$5.4 	imes 10^5 	ext{ V}$

Vedclass · Answer

(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$, where $k = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$.For point $A$, the distance from $+q$ is $x = 2 	ext{ cm} = 0.02 	ext{ m}$ and the distance from $-q$ is $(x+y) = 2+3 = 5 	ext{ cm} = 0.05 	ext{ m}$.$V_A = \frac{kq}{x} + \frac{k(-q)}{x+y} = kq \left( \frac{1}{0.02} - \frac{1}{0.05} ight) = (9 	imes 10^9)(10^{-6}) (50 - 20) = 9 	imes 10^3 	imes 30 = 2.7 	imes 10^5 	ext{ V}$.For point $B$, the distance from $+q$ is $(x+y) = 5 	ext{ cm} = 0.05 	ext{ m}$ and the distance from $-q$ is $x = 2 	ext{ cm} = 0.02 	ext{ m}$.$V_B = \frac{kq}{x+y} + \frac{k(-q)}{x} = kq \left( \frac{1}{0.05} - \frac{1}{0.02} ight) = (9 	imes 10^9)(10^{-6}) (20 - 50) = 9 	imes 10^3 	imes (-30) = -2.7 	imes 10^5 	ext{ V}$.The potential difference is $V_A - V_B = 2.7 	imes 10^5 - (-2.7 	imes 10^5) = 5.4 	imes 10^5 	ext{ V}$.

The potential difference $(V_A-V_B)$ in the arrangement shown in the figure is $(q=1 \mu C, x=2 \text{ cm}, y=3 \text{ cm})$.

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