A conducting rod PQ of length L = 1.0m is moving with a uniform speed v = 2.0 `m//s` in a uniform B magnetic field B = 4.0 T directed into the paper. A capacitor of capacity `C = 10 mu F` is connected as shown in the figure. Then what are the charges on the plates A and B of the capacitor

Explanation

The motional emf is e = Blv
`:.` p, d across the capacitor = Blv = `4 xx 1 xx 2 = 8V`
`q=CV= 10 xx 8 = 80 mu C`.
A is + Ve w.r.t B (from fleming right hand rule)
The charge on plate A is `q_(A) = +8O mu c`
The charge on plate B is `q_(B) = -80 mu C`