Question

(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the battery? V (b) If the same capacitor is connected to another battery and 192.0 µC of charge is stored on the capacitor, what is the voltage of the battery?

Answer 1

Answer:

a.6 V

b.24 V

Explanation:

We are given that

a.$C=8\mu F=8\times 10^{-6} F$

$1\mu =10^{-6}$

Q=$48\mu C=48\times 10^{-6} C$

We know that

$V=\frac{Q}{C}$

Using the formula

$V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V$

b.$Q=192\mu C=192\times 10^{-6} C$

$V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V$