A conducting rod of length `L=0.1 m` is moving with a uniform sped `v=0.2m//s` on conducting raisli a magnetic field `B=0.5T` as shown. On one side, the end of the rails is connected to a capacitor of capacitance `C=20muF`. Then, the charge on the capacitor's plates are

Options

A. `q_A=0=q_B`
B. `q_A=+20muC and q_B=-20muC`
C. `q_A=+0.2muC and q_B=-0.2muC`
D. `q_A=-0.2C and q_B=-0.2muC`

Answer

C

Explanation

From right hand rule, we can see that
`V_AgtV_B`
`:. q_A` is positive and `q_B ` negative
`q=CV=C=(Bvl)`
`=(20xx10^-6)(0.5)(0.1)`
`=0.2xx10^-6C=0.2muC`