Question

2 A parallel-plate capacitor has a charge of 6.0 pc when charged by a potential difference of 1.25 V. a. Find its capacitance. b. How much electrical potential energy is stored when this capacitors connected to a 1.50 V battery? itor hasanacitance of 200 nF

Answer 1

(a)

Given data

*The given charge is

`q=6.0\text{ pc = 6.0}*10^(-12)\text{ C}`

*The given potential difference is V = 1.25 V

The formula for the capacitance is given as

`C=(q)/(V)`

Substitute the known values in the above expression as

`\begin{gathered} C=(6.0*10^(-12))/(1.25) \\ =4.8*10^(-12)\text{ V} \end{gathered}`

Hence, the capacitance is

`C=4.8*10^(-12)\text{ V}`

(b)

Given data

*The given battery voltage is V = 1.50 V

*The given capacitance is

`C=200\text{ nF = 200}*10^(-9)\text{ F}`

The expression for the electrical potential energy is given as

`U_p=(1)/(2)CV^2`

Substitute the known values in the above expression as

`\begin{gathered} U_p=(1)/(2)(200*10^(-9))(1.50)^2 \\ =2.25*10^(-7)\text{ J} \end{gathered}`

Hence, the electrical potential energy is

`U_p=2.25*10^(-7)\text{ J}`