1
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Question

# Charge Q is uniformly distributed over a ring of radius r .Electric field at a point on the axis is maximum at a distance x from the centre. x is equal to

A
r2
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B
r2
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C
2r
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D
r
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Solution

## The correct option is A r√2The electric field due to the ring on its axis at a distance x is given by:-E=kqx(x2+R2)3/2To find maximum electric field, we will use the concept of maximum and minimum :-dEdx=kq(x2+R2)3/2−3/2(x2+R2)1/2.2x2(x2+R2)Now, dEdx=0⇒(x2+R2)3/2=32×2x2(x2+R2)1/2⇒x2+R2=3x2⇒2x2=R2⇒x2=R22⇒x=±R√2

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