A parallel plate capacitor has plates of area 100 cm² and separation 0.25 mm. If this capacitor is charged with 25 pC, what is the average charge density on the positive plate?

Question

Quizplus · Accepted Answer

The formula for the capacitance of a parallel plate capacitor is:

$C=\frac{\epsilon_0A}{d}$

where $\epsilon_0$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the separation between the plates.

Substituting the given values, we get:

$C=\frac{(8.85	imes10^{-12}\ F/m) 	imes (100\ cm^2)}{(0.25\ mm)}=35.4	imes10^{-12}\ F$

The charge on the capacitor is given as 25 pC, which is equivalent to 25$	imes 10^{-12}$ C.

The electric field between the plates of the capacitor is given by:

$E=\frac{\sigma}{\epsilon_0}$

where $\sigma$ is the surface charge density on the plates.

The potential difference between the plates of the capacitor is given by:

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

where $Q$ is the charge on the capacitor.

The electric field E is uniform between the plates and is given by:

$E=\frac{V}{d}$

Equating the two expressions for E and solving for $\sigma$, we get:

$\sigma=\frac{Q}{A}=\epsilon_0\frac{V}{d}=\epsilon_0\frac{Q}{Cd}=\frac{C}{A}\frac{Q}{d}$

Substituting the given values, we get:

$\sigma=\frac{(35.4	imes10^{-12}\ F)}{(100\ cm^2)}\frac{(25	imes10^{-12}\ C)}{(0.25\ mm)}=8.85	imes10^{-9}\ C/m^2=8.85\ nC/m^2$

Since the question asks for the average charge density on the positive plate, we divide the surface charge density by 2 to get:

$\sigma_{avg}=\frac{8.85}{2}=4.43\ nC/m^2=4.43	imes10^{-9}\ C/m^2=4.43\ \mu C/m^2$

Hence, the answer is D) 2.5 nC/m$^2$.

A Parallel Plate Capacitor Has Plates of Area 100 Cm2

A Parallel Plate Capacitor Has Plates of Area 100 Cm²