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

a.6 V

b.24 V

Step-by-step explanation:

We are given that

a.
`C=8\mu F=8* 10^(-6) F`

`1\mu =10^(-6)`

Q=
`48\mu C=48* 10^(-6) C`

We know that

`V=(Q)/(C)`

Using the formula

`V=(48* 10^(-6))/(8* 10^(-6))=6 V`

b.
`Q=192\mu C=192* 10^(-6) C`

`V=(192* 10^(-6))/(8* 10^(-6))=24 V`