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.

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 \( (1/3)\rho rπ/\epsilon(R^2-r^2)\)

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TopNewestThat'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 \( (1/3)\rho rπ/\epsilon(R^2-r^2)\)

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Oh thanks a lot, and upvoted!

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@Spandan Senapati

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

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