Two electric charges 12,mu C and -6,mu C are | vedclass

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(C) Let the distance of point $P$ from the $-6\,\mu C$ charge be $x$ meters. The distance of point $P$ from the $12\,\mu C$ charge is $(0.2 + x)$ mete

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$0.20$

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(C) Let the distance of point $P$ from the $-6\,\mu C$ charge be $x$ meters. The distance of point $P$ from the $12\,\mu C$ charge is $(0.2 + x)$ meters.The electric potential $V$ at point $P$ due to both charges is given by:$V = \frac{1}{4\pi\varepsilon_0} \left[ \frac{-6 	imes 10^{-6}}{x} + \frac{12 	imes 10^{-6}}{0.2 + x} ight] = 0$Since $V = 0$,we have:$\frac{6 	imes 10^{-6}}{x} = \frac{12 	imes 10^{-6}}{0.2 + x}$Dividing both sides by $6 	imes 10^{-6}$:$\frac{1}{x} = \frac{2}{0.2 + x}$Cross-multiplying gives:$0.2 + x = 2x$$x = 0.2\, m$Thus,the distance of point $P$ from the $-6\,\mu C$ charge is $0.20\, m$.

Two electric charges $12\,\mu C$ and $-6\,\mu C$ are placed $20\, cm$ apart in air. There will be a point $P$ on the line joining these charges and outside the region between them,at which the electric potential is zero. The distance of $P$ from the $-6\,\mu C$ charge is.......$m$.

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