Two charges +q and -q,each 1 mu C,are arrange | vedclass

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(B) The potential at point $A$ due to $+q$ is $V_{A(+q)} = \frac{1}{4\pi\epsilon_0} \frac{q}{x}$ and due to $-q$ is $V_{A(-q)} = \frac{1}{4\pi\epsilon

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$5.4 	imes 10^5 	ext{ V}$

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(B) The potential at point $A$ due to $+q$ is $V_{A(+q)} = \frac{1}{4\pi\epsilon_0} \frac{q}{x}$ and due to $-q$ is $V_{A(-q)} = \frac{1}{4\pi\epsilon_0} \frac{-q}{x+y}$.Thus,$V_A = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{x} - \frac{1}{x+y} ight) = \frac{q}{4\pi\epsilon_0} \left( \frac{y}{x(x+y)} ight)$.Similarly,the potential at point $B$ is $V_B = \frac{1}{4\pi\epsilon_0} \frac{q}{x+y} + \frac{1}{4\pi\epsilon_0} \frac{-q}{x} = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{x+y} - \frac{1}{x} ight) = -V_A$.Therefore,$V_A - V_B = V_A - (-V_A) = 2V_A = 2 	imes \frac{q}{4\pi\epsilon_0} \left( \frac{y}{x(x+y)} ight)$.Given $q = 10^{-6} 	ext{ C}$,$x = 0.02 	ext{ m}$,$y = 0.03 	ext{ m}$,and $\frac{1}{4\pi\epsilon_0} = 9 	imes 10^9 	ext{ Nm}^2/	ext{C}^2$.$V_A - V_B = 2 	imes 9 	imes 10^9 	imes 10^{-6} 	imes \frac{0.03}{0.02(0.02+0.03)} = 18 	imes 10^3 	imes \frac{0.03}{0.02 	imes 0.05} = 18 	imes 10^3 	imes \frac{0.03}{0.001} = 18 	imes 10^3 	imes 30 = 5.4 	imes 10^5 	ext{ V}$.

Two charges $+q$ and $-q$,each $1 \mu C$,are arranged as shown in the figure. If $x=2 \text{ cm}$ and $y=3 \text{ cm}$,then the potential difference $(V_A - V_B)$ is:

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