Help please!

A ball of radius $R$ is uniformly charged with the volume density $\rho$ . Find the flux of the electric field strength vector across the ball’s section formed by the plane located at a distance $r < R$ from the centre of the ball.

#ElectricityAndMagnetism

Easy Math Editor

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Comments

That's easy......The q means that take a plane at a distance r from the centre and pass it through the sphere.The plane intercepts the sphere and forms a circular cross section.We need to find the flux of the electric field through this area.As usual move up a distance s and DX and rotate it to get a circular strip of area $dA= 2πxdx$ .Now $d(\phi)=EdAcos\alpha$ .Where $cos\alpha$ = $r/√(r^2+x^2)$ . $E=kQ/R^3*√r^2+x^2$ as its an internal point.So the req eq is $d(\phi)=kQ/R^3*√(r^2+x^2)*2πxdx*r/√(r^2+x^2)$ .Now integrate from $x=0$ to $x=√(R^2-r^2)$ .And $Q=\rho *4/3πR^3$ .To get the ans as $(1/3)\rho rπ/\epsilon(R^2-r^2)$

Spandan Senapati - 4 years, 2 months ago

Oh thanks a lot, and upvoted!

Harsh Shrivastava - 4 years, 2 months ago

@Spandan Senapati

Harsh Shrivastava - 4 years, 2 months ago

And upvote.it..ha ha....just joking...

Spandan Senapati - 4 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$