Question

2. A parallel-plate capacitor has an area of 2.0 cm², and the plates are separated by 2.0 mm. a. What is the capacitance? b. How much charge does this capacitor store when connected to a 6.0 V battery?

Answer 1

Given data:

* The area of the parallel plate capacitor is,

`\begin{gathered} A=2cm^2 \\ A=2*10^(-4)m^2^{} \end{gathered}`

* The distance between the plates is,

`\begin{gathered} d=2\text{ mm} \\ d=2*10^(-3)\text{ m} \end{gathered}`

Solution:

(a). The capacitance of the capacitor in terms of area and distance between the plates is,

`C=(\epsilon_(\circ)A)/(d)`
`\text{where }\epsilon_(\circ)\text{ is the electrical permittivity of the fr}ee\text{ spaces}`

Substituting the known values,

`\begin{gathered} C=8.85*10^(-12)*(2*10^(-4))/(2*10^(-3)) \\ C=8.85*10^(-13)\text{ F} \end{gathered}`

Thus, the value of the capacitanc is 8.85 times 10 power -13 Farad.

(b). The voltage across the battery is,

`V=6\text{ Volts}`

The charge stored in the capacitor in terms of the voltage and the capacitance is,

`\begin{gathered} C=(Q)/(V) \\ Q=CV \end{gathered}`

where Q is the charge stored in the capacitor

Substituting the known values,

`\begin{gathered} Q=8.85*10^(-13)*6 \\ Q=53.1*10^(-13)\text{ Coulomb} \end{gathered}`

Thus, the charge stored in the parallel plate capacitor is 53.1 times 10 power -13 coulomb.