Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20cm.

Two Large Parallel Conducting Plates Carrying Opposite Charges Of Equal Magnitude Are Separated By

  1. 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.
  2. Calculate the Potential Difference V that exists between the two conducting plates.
  3. 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 plates $1$, $2$ 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 two 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:

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

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

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