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Two point charges Q and –Q/4 are separated by a distance x. Then

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Electric field is zero at a point on the axis which is at a distance x on the right side of the charge –Q/4

Potential is zero at a point on the axis which is x/3 on the right side of the charge –Q/4

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

Potential is zero at a point on the axis which is x/5 on the left side of the charge –Q/4

answer is A, B, C.

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Detailed Solution

Let r be the distance on the right of −Q/4 where the potential is zero.

So,

$\frac{k Q}{(r + x)} - \frac{k Q}{4 r} = 0 \Rightarrow 4 r - r - x = 0 \Rightarrow r = x / 3$

When r be the distance on the left of −Q/4,

$\frac{k Q}{(x - r)} - \frac{k Q}{4 r} = 0 \Rightarrow 4 r - x + r = 0 \Rightarrow r = x / 5$

Let electric field is zero at a distance r on the right side of the charge Q/4, so

$\frac{k Q}{{(r + x)}^{2}} - \frac{k Q}{4 r^{2}} = 0 \Rightarrow 4 r^{2} - {(r + x)}^{2} = 0 \Rightarrow (2 r + r + x) (2 r - r - x) = 0$

∴2r+r+x=0⇒r=−x/3

2r−r−x=0⇒r=x

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