Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

A system consists of a thin charged wire ring of radius

R

and a very long uniformly charged thread oriented along the axis of the ring with one of its ends coinciding with the centre of the ring. The total charge of the ring is equal to

q

. The charge of the thread (per unit length) is equal to

λ

. Find the interaction force between the ring and the thread.

Accepted Answer

Given that the radius of ring is $R$ and it has charge $q$ along its axis. A thread having charge per unit length $λ$ is kept whose one end is at the centre of the charged ring.

We know that due to a charged ring, the electric field at a point on its axis situated at distance $x$ from the centre is given by,
$E = \frac{k q x}{{(R^{2} + x^{2})}^{\frac{3}{2}}}$

$(along line O E away from centre)$

On a small element $d x$ , the force due to electric field of ring is given by,

$\Rightarrow d F = E d q \Rightarrow d F = \frac{k q x}{{(R^{2} + x^{2})}^{\frac{3}{2}}} λ d x \Rightarrow F_{total} = \int d F = \int_{0}^{\infty} \frac{k q λ x}{{(R^{2} + x^{2})}^{\frac{3}{2}}} d x Put x = R \tan θ, d x = R \sec^{2} θ d θ Limit - 0 to \frac{π}{2}, \Rightarrow F_{total} = \int_{0}^{\frac{π}{2}} \frac{k q λ (R \tan θ) R \sec^{2} θ}{{(R^{2} + {(R \tan θ)}^{2})}^{\frac{3}{2}}} d θ \Rightarrow F_{total} = \frac{k q λ}{R} \int_{0}^{\frac{π}{2}} \frac{\tan θ \sec^{2} θ}{{(1 + \tan^{2} θ)}^{\frac{3}{2}}} d θ \Rightarrow F_{total} = \frac{k q λ}{R} \int_{0}^{\frac{π}{2}} \frac{\tan θ \sec^{2} θ}{\sec^{3} θ} d θ = \frac{k q λ}{R} \int_{0}^{\frac{π}{2}} \sin θ d θ \Rightarrow F_{total} = \frac{k q λ}{R} {|- \cos θ|}_{0}^{\frac{π}{2}} = \frac{k q λ}{R} \{- \cos (\frac{π}{2}) + \cos 0\} \Rightarrow F_{total} = \frac{k q λ}{R} = \frac{q λ}{4 π ε_{0} R}$

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