Question

20in the given circuit the charge on lower plate B is(1) Positive(2) Negative(3) Zero(4) Infinite​

Answer 1

The total charge on plate is 80μC80 \mu C80μC. If qBq_BqB​ and qCq_CqC​ charges on plates B and C then

qB+qC=80 ....(1)q_B+q_C=80  ....(1)qB​+qC​=80 ....(1)

As the capacitors B and C are in parallel so potential across both are equal .

i.e, qBCB=qCCC\dfrac{q_B}{C_B}=\dfrac{q_C}{C_C} CB​qB​​=CC​qC​​

or qB2=qC3\dfrac{q_B}{2}=\dfrac{q_C}{3}2qB​​=3qC​​

using (1),  80−qC2=qC3\dfrac{80-q_C}{2}=\dfrac{q_C}{3}280−qC​​=3qC​​

or  240−3qC=2qC⇒qC=48 μC240-3q_C=2q_C \Rightarrow q_C=48  \mu C240−3qC​=2qC​⇒qC​=48 μC

Now for sign of charge: as lower plate of C is connected to ground so upper plate of C should be positive.

Thus, charge on upper plate of 3μF3\mu F3μF is +48μC+48 \mu C+48μC

THIS IS THE ANSWER OF THE GIVIEN QUWSTION