CBSE Class 12 Physics 2016 Delhi Set 1 Paper

Question : 11

Total: 26

SECTION - C

A charge is distributed uniformly over a ring of radius ' a '. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.

Solution:

Obtaining an expression for electric field intensity
Showing behaviour at large distance

Net Electric Field at point P=

2πa

∫

dE‌cos‌θ
dE= Electric field due to a small element having charge dq
=‌

4πε0

‌

Let λ‌=‌ Linear charge density ‌
‌=‌

dq‌=λdl
Hence E=

2πa

∫

‌

4πε0

⋅‌

λdl

×‌

, where cos‌θ=‌

‌=‌

λx

4πε0r3

(2πa)
‌=‌

4πε0

‌

(x2+a2)‌

,
where total charge Q=λ×2πa
At large distance i.e., x>>a
E≃‌

4πε0

⋅‌

This is the electric field due to a point charge at distance x .

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