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Question

Physics Question on Electric Charge

Two point charges −q-q and +q/2+q/2 are situated at the origin and at the point (a,0,0)(a, 0, 0) respectively. The point along the XX-axis where the electric field vanishes is

A

x=a2x=\frac{a}{\sqrt{2}}

B

x=2ax=\sqrt{2a}

C

x=2a2−1x=\frac{\sqrt{2a}}{\sqrt{2-1}}

D

x=2a2+1x=\frac{\sqrt{2a}}{\sqrt{2+1}}

Answer

x=2a2−1x=\frac{\sqrt{2a}}{\sqrt{2-1}}

Explanation

Solution

Suppose the Field vanishes at a distance xx, we have Kqx2=Kq/2(x−a)2\frac{K q}{x^{2}}=\frac{K_{q / 2}}{(x-a)^{2}} or 2(x−a)2=x22(x-a)^{2}=x^{2} or 2(x−a)=x\sqrt{2}(x-a)=x or (2−1)x=2a(\sqrt{2}-1) x=\sqrt{2} a or x=2a2−1x=\frac{\sqrt{2} a}{\sqrt{2}-1}