**Calculate the absolute magnitude of Electric Field E in the area between the two conducting plates if the magnitude of charge density at the surface of each place is 47.0 nC/m^2.****Calculate the Potential Difference V that exists between the two conducting plates.****Calculate the impact on the magnitude of Electric Field E and Potential Difference V if the distance between the conducting plates is doubled while keeping the density of charge constant at the conducting surfaces.**

The aim of this article is to find the **Electric Field** $\vec{E}$ and **Potential Difference** $V$ between the **two conducting plates** and the impact of change in the distance between them.

The main concept behind this article is **Electric Field** $\vec{E}$ and **Potential Difference** $V$.

**Electric Field** $\vec{E}$ acting on a plate is defined as the **electrostatic force** in terms of unit charge that act on a unit area of the plate. It is represented by **Gauss Law** as follows:

\[\vec{E}=\frac{\sigma}{2\in_o}\]

Where:

$\vec{E}=$ **Electric Field**

$\sigma=$ **Surface Charge Density of the Surface**

$\in_o=$ **Vacuum Permittivity** $= 8.854\times{10}^{-12}\dfrac{F}{m}$

**Potential Difference** $V$ between two plates is defined as the **electrostatic potential energy** in terms of unit charge that acts between those two plates separated by a certain distance. It is represented as follows:

\[V=\vec{E}.d\]

Where:

$V=$ **Potential Difference**

$\vec{E}=$ **Electric Field**

$d=$ **Distance between two plates**

## Expert Answer

Given that:

**Distance between two plates** $d=2.2cm=2.2\times{10}^{-2}m$

**Surface Charge Density of each plate** $\sigma=47.0\dfrac{n.C}{m^2}=47\times{10}^{-9}\dfrac{C}{m^2}$

**Vacuum Permittivity** $\in_o=8.854\times{10}^{-12}\dfrac{F}{m}$

**Part (a)**

**Magnitude of Electric Field** $\vec{E}$ acting between given two **parallel plates** $1$, $2$ is:

\[\vec{E}={\vec{E}}_1+{\vec{E}}_2\]

\[\vec{E}=\frac{\sigma}{2\in_o}+\frac{\sigma}{2\in_o}\]

\[\vec{E}=\frac{2\sigma}{2\in_o}=\frac{\sigma}{\in_o}\]

Substituting the value of **Surface Charge Density** $\sigma$ and **Vacuum Permittivity** $\in_o$:

\[\vec{E}=\frac{47\times{10}^{-9}\dfrac{C}{m^2}}{8.854\times{10}^{-12}\dfrac{F}{m}}\]

\[\vec{E}=5.30834\times{10}^3\frac{N}{C}\]

\[Electric\ Field\ \vec{E}=5308.34\frac{N}{C}=5308.34\frac{V}{m}\]

**Part (b)**

**Potential Difference** $V$ between given **two parallel plate**s $1$, $2$ is:

\[V=\vec{E}.d\]

Substituting the value of **Electric Field** $\vec{E}$ and the **distance** $d$ between two plates, we get:

\[V=5.30834\times{10}^3\frac{V}{m}\times2.2\times{10}^{-2}m\]

\[Potential\ Difference\ V=116.78\ V\]

**Part (c)**

Given that:

The **distance** between the t**wo parallel plates** is **double**.

As per the expression of **Electric Field** $\vec{E}$, it is not dependent on distance, hence any change in distance between the parallel plates will not have any impact on **Electric Field** $\vec{E}$.

\[\vec{E}=5308.34\frac{V}{m}\]

We know that the **Potential Difference** $V$ between given two **parallel plates** $1$, $2$ is:

\[V=\vec{E}.d\]

If the **distance** is **doubled**, then:

\[V^\prime=\vec{E}.2d=2(\vec{E}.d)=2V\]

\[V^\prime=2(116.78\ V)=233.6V\]

## Numerical Result

**Part (a) – Magnitude of Total Electric Field** $\vec{E}$ acting between given **two parallel plates** $1$, $2$ will be:

\[Electric\ Field\ \vec{E}=5308.34\frac{N}{C}=5308.34\frac{V}{m}\]

**Part (b) – Potential Difference** $V$ between given **two parallel plates** $1$, $2$ is:

\[V=116.78\ V\]

**Part (c)** – If the **distance** between the conducting plates is **doubled**, **Electric Field** $\vec{E}$ will not change whereas the **Potential Difference** $V$ will be **doubled**.

## Example

Calculate the magnitude of **Electric Field** $\vec{E}$ in the area between the **two conducting plates** if the **surface charge density** of each place is $50\dfrac{\mu C}{m^2}$.

**Solution**

**Magnitude of Total Electric Field** $\vec{E}$ acting between given **two parallel plates** $1$, $2$ will be:

\[\vec{E}={\vec{E}}_1+{\vec{E}}_2\]

\[\vec{E}=\frac{\sigma}{2\in_o}+\frac{\sigma}{2\in_o}=\frac{\sigma}{\in_o}\]

Substituting the values, we get:

\[\vec{E}=\frac{50\times{10}^{-6}\dfrac{C}{m^2}}{8.85\times{10}^{-12}\dfrac{F}{m}}\]

\[\vec{E}=5.647\times{10}^6\frac{N}{C}=5.647\times{10}^6\frac{V}{m}\]