Question

A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positive plate of each connected to the negative plate of the other. What is the final charge on the 6.0-mF capacitor?

Answer 1

Answer:

12mC

Explanation:

We are given that

$C_1=4 mF$

$V_1=50 V$

$C_2=6mF$

$V_2=30 V$

We have to find the final charge on the 6mF capacitor.

We know that

$Q=CV$

Using the formula

$Q_1=4\times 50=200mC$

$Q_2=6\times 30=180mC$

Total charge=$Q_1-Q_2=200-180=20mC$

Let V be the final potential

$4V+6V=20mC$

$10V=20mC$

$V=\frac{20}{10}=2 V$

Final charge,q=CV=$6\times 2=12mC$

Hence, the final charge on the 6.0 mF capacitor=12mC