Question

In Fig., let C1 = 3.00 mF, C2 = 5.00 mF, and Vab = +64.0 V. Calculate

(a) the charge on each capacitor and

(b) the potential difference across each capacitor.

Short answer

a) The charge on each capacitor is $120.0\mu C$.

b) The potential difference for the first capacitor is 40.0V

The potential difference for the first capacitor is 24.0V

Step-by-step solution

Step 1:

We are given $C_1\&C_2$are connected in series where $C_1=3.0pF\&C_2=5.0pF$the potential between a and b is $V_{ab}=64.0V$

(a) We will find that the total charge on the lower plate of Ci and the upper plate of C, together must always be zero because these plates are not connected to anything except each other. Thus in a series connection, the magnitude of the charge on all plates is the same and the charge on the first capacitor is the same for the second capacitor and it will be given by

$Q=C_{eq}{V}_{ab}$

Where is the equivalent capacitance of , and , as they are in series and it is given by

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2} \\ \frac{1}{C_{eq}}=\frac{1}{3.00\mu F}+\frac{1}{5.00\mu F} \\ \frac{1}{C_{eq}}=0.533 \\ \frac{1}{C_{eq}}=1.875\mu F \end{gathered}$$

Substitute values in the above equation

$Q=C_{eq}{V}_{ab}=\left(1.875\mu F\right)\left(64.0V\right)=120.0\mu C$

Hence, the charge on each capacitor is$120.0\mu C$

Step 2:

(b) In a series connection, the potential differences of the individual capacitors are not the same unless their capacitances are the same, so we can calculate the potential difference for each capacitor by dividing the charge Q by the capacitance of each capacitor as next:

For the first capacitor, the potential difference is and given by

$V_1=\frac{Q}{C_{1-}}=\frac{120.0\mu C}{3.0\mu F}=40.0V$

Hence, the potential difference for the first capacitor is 40.0V

For the first capacitor, the potential difference isand given by

$V_2=\frac{Q}{C_2}=\frac{120.0\mu C}{5.0\mu F}=24.0V$

Hence, the potential difference for the first capacitor is 24.0V