Question

written part:
write a list of information given, choose a right equation, substitute information into your equations,
solve a problem mathematically, and do not forget units in your answer.

1. a parallel - plate capacitor has plates of area ( 0.40 , m^{2} ) and plate separation of ( 0.20 , mm ). the

capacitor is connected across a ( 9.0 - v ) potential source. ( (epsilon_{0}=8.85\times10^{-12}c^{2}/(ncdot m^{2})) )
(a) what is the magnitude of the electric field between the plates?
(b) what is the capacitance of the capacitor?
(c) what is the magnitude of the charge on each plate of the capacitor?

Explanation:

Step1: Find the electric field (a)

The formula for the electric field \(E\) between the plates of a parallel - plate capacitor connected to a potential source \(V\) with plate separation \(d\) is \(E=\frac{V}{d}\).
Given \(V = 9.0\space V\) and \(d=0.20\times10^{-3}\space m\).
Substitute the values: \(E=\frac{9.0}{0.20\times 10^{-3}}\)

$$E = 4.5\times10^{4}\space V/m$$

Step2: Calculate the capacitance (b)

The formula for the capacitance \(C\) of a parallel - plate capacitor is \(C=\frac{\epsilon_{0}A}{d}\), where \(\epsilon_{0}=8.85\times 10^{-12}\space C^{2}/(N\cdot m^{2})\), \(A = 0.40\space m^{2}\) and \(d=0.20\times10^{-3}\space m\)
Substitute the values: \(C=\frac{8.85\times 10^{-12}\times0.40}{0.20\times10^{-3}}\)

$$C = 1.77\times10^{-8}\space F$$

Step3: Determine the charge (c)

The formula for the charge \(Q\) on a capacitor is \(Q = CV\).
We know \(C = 1.77\times10^{-8}\space F\) and \(V = 9.0\space V\)
Substitute the values: \(Q=(1.77\times10^{-8})\times9.0\)

$$Q = 1.593\times10^{-7}\space C$$

Answer:

(a) \(4.5\times10^{4}\space V/m\)
(b) \(1.77\times10^{-8}\space F\)
(c) \(1.593\times10^{-7}\space C\)