Question

Suppose parallel metal plates are 0.50 cm apart and are connected to a 90 V battery. What is the surface charge density of plates?

Answer 1

Step 1: Given data

1. The distance between the plates is $d=0.50cm=0.005m.$
2. The supplied voltage between the plates is $V=90volts.$

Step 2: Formulae used

1. We know that for parallel plate capacitors the electric field due to charge flow is $E=\frac{\sigma }{\epsilon }$, where, $\sigma$ is the surface charge density of the capacitor and $\epsilon$ is the permittivity of the space between the metal plates.
2. Again we know that the electric field $\mathrm{E}$ of a conductor of length $\mathrm{L}$ is associated with a voltage $\mathrm{V}$ that is $E=\frac{V}{L}$.
3. The permittivity of air is $8.854\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2}$

Step 3: Diagram

Step 4: Find the surface charge density

Considering the space between the plates is filled with air. As we know, the electric field in a parallel plate capacitor is $E=\frac{\sigma }{\epsilon }$.

So, the surface charge density is:

$$\begin{gathered} \sigma =\epsilon E \\ \sigma =\epsilon \frac{V}{d}=\left(8.854\times {10}^{-12}\right)\times \left(\frac{90}{0.005}\right)\left(\sin ce,E=\frac{V}{d}\right) \\ \sigma =1.6\times {10}^{-7}C/m^2 \end{gathered}$$

Therefore, the surface charge density of plates is $1.6\times {10}^{-7}C/m^2$.