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A semicircular frame of radius R is having a charge q and -q in respective quadrants (as shown in the image). The frame is able to rotate about the point C in uniform electric field E. The work done to rotate the frame so the base AB if semicircle becomes parallel to electric field is?

A semicircular frame of radius R is having a charge q and -q in respective quadrants (as shown in the image). The frame is able to rotate about the point C in uniform electric field E. The work done to rotate the frame so the base AB if semicircle becomes parallel to electric field is?

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1 Answers

Eshan
askIITians Faculty 2095 Points
5 years ago

568-247_charge.png
Since all positive charges have moved to a position vertically below them, they have displaced perpendicular to the electric field, and thus no work has been done in moving them.

However, negative charges have been displaced against the electric force.
Work done in moving a small charge isdW=F.s=(dq)E.s=(\lambda Rd\theta)E.(Rsin\theta)
Hence the total work done=\int_0^{\pi/2}(\lambda Rd\theta)E.(Rsin\theta)
where\lambda =\dfrac{q}{\pi R/2}
\lambda R^2E\int_0^{\pi/2}sin\theta d\theta =\lambda R^2E
=(\dfrac{q}{\pi R/2})R^2E
=\dfrac{2R}{\pi}qE

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