Find the Electric field!

Electricity and Magnetism Level 2

An infinitely large surface of uniform charge density σ \sigma has a disc of radius R R cut out (see figure). The magnitude of the electric field at a distance a a above the center of the disc is given by?

σ π a 2 ϵ 0 R 2 + π 2 a 2 \large \frac{\sigma \pi a}{2 \epsilon_0 \sqrt{R^2 + \pi^2 a^2}} σ a 2 2 ϵ 0 ( R 2 + a 2 ) \large \frac{\sigma a^2}{2 \epsilon_0 (R^2 + a^2)} σ a ϵ 0 R 2 + 4 a 2 \large \frac{\sigma a}{\epsilon_0 \sqrt{R^2 + 4a^2}} σ a 2 ϵ 0 R 2 + a 2 \large \frac{\sigma a}{2 \epsilon_0 \sqrt{R^2 + a^2}}

Nishant Rai by Nishant Rai , 25, India

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2 solutions

Nishant Rai
May 27, 2015

Note : For finding the electric field at a distance x x from a charged circular disc of surface charge density σ \sigma , one should consider elementary rings of charges on the disc and then integrate.

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Shit man. I by mistake took it a ring. Anyways thanx

Kyle Finch - 6 years ago

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It happens sometimes. ⌣ ¨ \ddot \smile

@Kyle Finch

Nishant Rai - 6 years ago
Rwit Panda
Jun 22, 2015

Consider elemental ring of radius x and thickness dx on sheet. Using this calculate the net electric field by putting limits R to infinity.

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