Two electric charges of 9 mu C and -3 mu C ar | vedclass

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(B) Given: $q_1 = 9 \mu C$,$q_2 = -3 \mu C$,and the distance $r = 0.16 \ m$.Let the distance of point $P$ from $q_1$ be $x$. Then the distance of $P$

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$0.12$

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(B) Given: $q_1 = 9 \mu C$,$q_2 = -3 \mu C$,and the distance $r = 0.16 \ m$.Let the distance of point $P$ from $q_1$ be $x$. Then the distance of $P$ from $q_2$ is $(0.16 - x)$.The electric potential $V$ at point $P$ due to both charges is the sum of individual potentials:$V = V_1 + V_2 = 0$$\frac{1}{4 \pi \epsilon_0} \frac{q_1}{x} + \frac{1}{4 \pi \epsilon_0} \frac{q_2}{0.16 - x} = 0$$\frac{9 	imes 10^{-6}}{x} + \frac{-3 	imes 10^{-6}}{0.16 - x} = 0$$\frac{9}{x} = \frac{3}{0.16 - x}$$3(0.16 - x) = x$$0.48 - 3x = x$$4x = 0.48$$x = 0.12 \ m$Thus,the distance of point $P$ from the $9 \mu C$ charge is $0.12 \ m$.

Two electric charges of $9 \mu C$ and $-3 \mu C$ are placed $0.16 \ m$ apart in air. There will be a point $P$ on the line joining the two charges and in between them where the electric potential is zero. The distance of $P$ from the $9 \mu C$ charge is (in $m$)

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