Question

In Fig., let C1 = 3.00 mF, C2 = 5.00 mF, and Vab = +52.0 V. Calculate

(a) the charge on each capacitor and

(b) the potential difference across each capacitor.

Short answer

The potential difference is the same for the capacitor and equals the potential between a and b and that is 52.0V

The charge on the capacitor $C_1$is $156.0\mu C$.

The charge on the capacitor$C_1\mathrm{is}156.0\mu C$

Step-by-step solution

Potential difference and charge on each Capacitor.

We are given$C_1$and$C_2$are connected in parallel where $C_1$=3.00 uF and $C_2$= 5.00 pF and the potential between a and b is $V_{ab}$= 520 V

(a) In Figure 24.9a, the two capacitors are connected in parallel. As shown in the

figure the upper plates of the two capacitors are connected by conducting wires

to form an equipotential surface, and the lower plates form another. Therefore, in a

parallel connection, the potential difference for all individual capacitors is the

same and is equal to

$V_{ab}=V_1=V_2=52.0V$

So we can get the charge for each capacitor by using the value of the given

potential and the capacitance of each capacitor as next

For the first capacitor $C_1$, the charge$Q_1$is given by

$Q_1=C_1{V}_{ab}=\left(3.00\mu F\right)\left(52.0V\right)=156.0\mu C$

For the second capacitor $C_2$, the charge$Q_2$is given by

$Q_2=C_2{V}_{ab}=\left(5.00\mu F\right)\left(52.0V\right)=260.0\mu C$

Conclusion

Hence, the potential difference is the same for the capacitor

and equals the potential between a and b.

$V_{ab}=V_1=V_2=52.0V$