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## QuestionPhysicsClass 12

Two charges $$3\times 10^{-8}\;\text C$$ and $$-2\times 10^{-8}\;\text C$$  are located $$15\;\text{cm}$$  apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Let us take the origin $$O$$  at the location of the positive charge. The line joining the two charges is taken to be the $${x-}$$axis ; the negative charge is taken to be on the right side of the origin.
Let $$P$$  be the required point on the $${x -}$$axis  where the potentlal is zero.
If $$x$$  is the $${x -}$$coordinate of $$P,$$  obviously $$x$$ must be positive. $$($$There is no possibility of potential due to the two charges adding up to zero for $$x<0)$$ If $$x$$  lies between $$O$$  and $$A,$$  we have
$$\frac 1{4\pi \varepsilon _0}\left[\frac{3\times 10^{-6}}{x\times 10^{-2}}-\frac{2\times 10^{-8}}{\left(15-x\right)\times 10^{-2}}\right]=0$$
where $$x$$  is in $$\text{cm}$$ . That is.
$$\frac 3 x-\frac 2{15-x}=0$$
which gives $$x=9\;\text{cm}$$
If $$x$$  lies on the extended line $$\mathit{OA}$$ , the required condition is
$$\frac 3 x-\frac 2{x-15}=0$$
which gives   $$x=45\;\text{cm}$$
Thus, electric potential is zero at $$9\;\text{cm}$$  and $$45\;\text{cm}$$  away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.