Question

A 4.0-mF capacitor initially charged to 50 V and 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 of the 6.0-mF capacitor?

a. 20 mC
b. 8.0 mC
c. 10 mC
d. 12 mC
e. 230 mC

Answer 1

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12 mC

Explanation:

We are given that...

$C_{1} = 4mF \\ V_{1} = 50v \\ C_{2} = 6mF \\ V_{2} = 30v \\ \\ we've \: to \: find \: the \: final \: charge \: on \: the \: 6mF \\capacitor \:- \\ \\ we \: know \: that ,\: \\ Q = CV \\ \\ using \: the \: formula \: \\ Q_{1} = 4 \times 50 = 200mC \\ Q_{2} = 6 \times 30 = 180mC \\ \\ total \: charge , \: Q_{1} - Q_{2} = 200 - 180 = 20mC \\ \\ let \: V \: be \: the \: final \: potential \: \\ 4V + 6V = 20mC\\ 10V = 20mC \\ V = \frac{20}{10} = 2V \\ final \: charge \: Q = CV = 6 \times 2 = 12mC$

therefore,

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