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Question

The electric field at a point P at a distance x from the centre of the ring is
591280.JPG
  1. kQ(x2+R2)3/2
  2. kQx(x2+R2)1/2
  3. kQ(x2+R2)1/2
  4. kQx(x2+R2)3/2

A
kQ(x2+R2)3/2
B
kQx(x2+R2)1/2
C
kQ(x2+R2)1/2
D
kQx(x2+R2)3/2
Solution
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Let us consider a small charge element of charge dq

And, dq=Q2πR.

The field at point P due to this element is =E=Kdqr2

E=Kdq(R2+x2)

Now, from figure we see that component of field normal to axis is cancelled by two diametrically opposite points.

Hence, only component of field along axis is left which add up for all such elements.

Enet=Ecosθ where θ is same for all elements means θ=constant

Enet=Kcosθ(R2+x2)dq

Enet=KQ(R2+x2)cosθ

Enet=KQ(R2+x2)xR2+x2

Enet=KQx(R2+x2)3/2

Answer-(D)

846659_591281_ans_78badedd65bd453ab04099190718957d.jpg

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