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Two point charges Q and $- Q / 4$ placed along x-axis are separated by a distance r. Take $- Q / 4$ as origin and it is placed right of Q. Then, potential is zero:

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Both at $x = r / 3$ and at $x = - r / 5$ .

At $x = - r / 5$ only.

There exist two points on the axis where electric field is zero.

At $x = r / 3$ only.

answer is C.

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Detailed Solution

$\frac{k Q}{4 x_{1}} = \frac{k Q}{r + x_{1}}$

$r + x_{1} = 4 x_{1}$

$x_{1} = \frac{r}{3}$

$\frac{k Q}{4 x_{2}} = \frac{k Q}{r - x_{2}}$

$x_{2} = \frac{r}{5}$

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