The electric field at a point P at a distance x from the centre of the ring is
kQ(x2+R2)3/2
kQx(x2+R2)1/2
kQ(x2+R2)1/2
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
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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)x√R2+x2
⟹Enet=KQx(R2+x2)3/2
Answer-(D)
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