Question
Question: In the given circuit, a charge of \(+80\mu C\) is given to the upper plate of the \(4\mu F\) capacit...
In the given circuit, a charge of +80μC is given to the upper plate of the 4μF capacitor. Then at the steady state, the charge on the upper plate of the 3μF capacitor will be given as,
A.+32μCB.+40μCC.+48μCD.+80μC
Solution
The capacitors, B and C are parallel, therefore the potential across both the capacitors will be equal. This relation is to be used in the equation of total charge. Rearranging this equation will give the answer. As the lower plate of C is connected to ground so the upper plate of C will be positive. These all may help you in solving the question.
Complete answer:
As per the question, we know that the total charge on the plate will be given as 80μC. Let us assume that the charge on the plate B be qB and charge on the plate C be qC. Therefore we can write that,
qC+qB=80μC
As we can see in the circuit that the capacitors B and C are parallel. Therefore the potential across both these capacitors will be identical. That is we can write that,
CBqB=CCqC
This on the basis of the equation of potential given as,
V=Cq
Substitute the value of capacitance in the equation,
2qB=3qC⇒qB=32qC
We can substitute this in the equation of total charge. That is,
qC+qB=80μCqC+32qC=80μC
Rearranging the equation will give,
280−qC=3qC
Simplifying this equation will give,
240−3qC=2qC
That is,
qC=48μC
In the case of the sign of charge, as it is shown that the lower plate of C is connected to ground therefore the upper plate of C should be positive.
Hence the charge on the upper plate of 3μF is +48μC. Hence the answer for the question has been calculated.
The answer is option C.
Note:
The capacitor is a device used to store the electrical energy in general. The efficiency of the capacitor in performing its function is mentioned using the quantity called capacitance. The charge has been stored using two metal plates separated by a small distance.