Electric field intensity due to a uniformly charged infinite plane thin sheet:
Let us consider, A plane charged sheet (It is a thin sheet so it will have
surface charge distribution whether it is a conducting or nonconducting
sheet) whose surface charge density is $\sigma$. From symmetry, Electric
field intensity is perpendicular to the plane everywhere and the field
intensity must have the same magnitude on both sides of the sheet. Let point
$P_{1}$ and $P_{2}$ be the two-point on the opposite side of the sheet.
To use Gaussian law, we construct a cylindrical Gaussian surface of
cross-section area $\overrightarrow{dA}$, which cuts the sheet, with points
$P_{1}$ and $P_{2}$. The electric field $\overrightarrow{E}$ is normal to
end faces and is away from the plane. Electric field $\overrightarrow{E}$ is
parallel to cross-section area $\overrightarrow{dA}$. Therefore the curved
cylindrical surface does not contribute to the flux i.e. $\oint
\overrightarrow{E} \cdot \overrightarrow{dA}=0$.Hence the total flux is
equal to the sum of the contribution from the two end faces. Thus, we get
$ \phi_{E}=\int_{A} \overrightarrow{E} \cdot
\overrightarrow{dA}+\int_{A} \overrightarrow{E} \cdot \overrightarrow{dA}$
$ \phi_{E}= \int_{A} E \: dA \:cos 0^{\circ} +\int_{A} E \: dA \:cos
0^{\circ} $
Here the direction of $\overrightarrow{E}$ and $\overrightarrow{dA}$ is same. So the angle will be $\theta = 0^{\circ}$.
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| Infinite plane thin sheet |
$ \phi_{E}= \int_{A} E \: dA +\int_{A} E \: dA $
$ \phi_{E}= \int_{A} 2E \: dA $
$ \phi_{E}= 2E \int_{A} \: dA $
$ \phi_{E}= 2E\:A $
$ \frac{q}{\epsilon_{0}}=2E\:A \qquad \left \{\because
\phi_{E}=\frac{q}{\epsilon_{0}} \right \}$
$ E=\frac{q}{2\epsilon_{0} A}$
$\because q=\sigma A $, So the above equation can be written as:
$ E=\frac{\sigma A}{2\epsilon_{0} A} $
$ E=\frac{\sigma}{2\epsilon_{0}} $
Electric field intensity due to the uniformly charged infinite conducting
plane thick sheet or Plate:
Let us consider that a large positively charged plane sheet having a finite
thickness is placed in the vacuum or air. Since it is a conducting plate so the
charge will be distributed uniformly on the surface of the plate. Let
$\sigma$ be the surface charge density of the charge
Let's take a point $P$ close to the plate at which electric field intensity
has to determine. Since there is no charge inside the conducting plate, this
conducting plate can be assumed as equivalent to two plane sheets of charge
i.e sheet 1 and sheet 2.
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| Plane Charged Plate |
The magnitude of the electric field intensity $\overrightarrow {E_{1}}$ at
point $P$ due to sheet 1 is →
$ E_{1}=\frac{\sigma}{2\epsilon_{0}}$ (away from sheet 1)
The magnitude of the electric field intensity $\overrightarrow {E_{2}}$ at
point $P$ due to sheet 2 is →
$ E_{2}=\frac{\sigma}{2\epsilon_{0}}$ (away from sheet 2)
Since $\overrightarrow {E_{1}}$ and $\overrightarrow {E_{2}}$ are in the
same direction, the magnitude of resultant intensity $\overrightarrow {E}$
at point $P$ due to both the sheet is →
$ E=E_{1}+E_{2}$
$\because \quad E=\frac{\sigma}{2\epsilon_{0}}+\frac{\sigma}{2\epsilon_{0}}$
$ E=\frac{\sigma}{\epsilon_{0}} $
The resultant electric field will be away from the plate. If the plate is
negatively charged, the electric field intensity $\overrightarrow {E}$ would
be directed toward the plate.
We have obtained the above formula for a 'plane' charged conductor. In fact,
it holds for the electric field intensity 'just' outside a charged conductor
of any shape.
Electric field intensity due to two Infinite Parallel Charged Sheets:
When both sheets are positively charged:
Let us consider, Two infinite, plane, sheets of positive charge, 1 and 2 are
placed parallel to each other in the vacuum or air. Let $\sigma_{1}$ and
$\sigma_{2}$ be the surface charge densities of charge on sheet 1 and 2
respectively.
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| Likely positive charged sheet |
Let $\overrightarrow {E_{1}}$ and $\overrightarrow {E_{2}}$ be the electric
field intensities at any point due to sheet 1 and sheet 2 respectively.
Then,
The electric field intensity at points $P'$ →
$ E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$ (away from sheet 1)
$ E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$ (away from sheet 2)
Since, Electric field intensities $\overrightarrow {E_{1}}$ and
$\overrightarrow {E_{2}}$ are in the same direction, the magnitude of
resultant intensity at point $P'$ is given by →
$ E=E_{1}+E_{2}$
$ E=\frac{\sigma_{1}}{2 \epsilon_{0}}+\frac{\sigma_{2}}{2
\epsilon_{0}}$
$ E=\frac{1}{2 \epsilon_{0}} \left (\sigma_{1}+\sigma_{2} \right )$
If both sheets have equal charge densities $\sigma$ i.e.
$\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:
$ E=\frac{\sigma}{ \epsilon_{0}} $
This electric field intensity would be away from both sheet 1 and sheet 2.
The electric field intensity at points $P$→
Electric field intensity at point $P$ due to sheet 1 is →
$ E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$ (away from sheet 1)
$ E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$ (away from sheet 2)
Now, both electric field intensities $\overrightarrow{E_{1}}$ and
$\overrightarrow{E_{2}}$ are in opposite direction. The magnitude of
resultant electric field $\overrightarrow{E}$ at point $P$ is given by
$ E= E_{1}-E_{2}$
$ E=\frac{\sigma_{1}}{2 \epsilon_{0}}-\frac{\sigma_{2}}{2
\epsilon_{0}}$
$ E=\frac{1}{2 \epsilon_{0}} \left (\sigma_{1}-\sigma_{2} \right )$
If both sheets have equal charge densities $\sigma$ i.e.
$\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:
$ E=0 $
The electric field intensity at points $P''$ →
$ E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$ (away from sheet 1)
$ E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$ (away sheet 2)
Since, Electric field intensities $\overrightarrow {E_{1}}$ and
$\overrightarrow {E_{2}}$ are in the same direction, the magnitude of
resultant intensity at point $P''$ is given by →
$ E=E_{1}+E_{2}$
$ E=\frac{\sigma_{1}}{2 \epsilon_{0}}+\frac{\sigma_{2}}{2
\epsilon_{0}}$
$ E=\frac{1}{2 \epsilon_{0}} \left (\sigma_{1}+\sigma_{2} \right )$
If both sheets have equal charge densities $\sigma$ i.e.
$\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:
$ E=\frac{\sigma}{ \epsilon_{0}} $
This electric field intensity would be away from both sheet 1 and sheet 2.
When one-sheet is positively charged and the other sheet negatively charged:
Let us consider two sheets 1 and 2 of positive and negative charge densities
$\sigma_{1}$ and $\sigma_{2}$ ($\sigma_{1} > \sigma_{2}$)
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| Unlike charged parallel Sheet |
The electric field intensities at point $P'$ →
$ E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$ (away from sheet 1)
$ E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$ (toward from sheet 2)
The magnitude of the resultant electric field $E$
$ E=E_{1}-E_{2}$
$E= \frac{1}{2\epsilon_{0}} \left ( \sigma_{1}- \sigma_{2}\right )$
If both sheets have equal charge densities $\sigma$ i.e.
$\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:
$ E=0 $
The electric field intensities at point $P$ →
$ E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$ (away from sheet 1)
$ E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$ (towards sheet 2)
Since, Electric field intensities $\overrightarrow {E_{1}}$ and
$\overrightarrow {E_{2}}$ are in the same direction, the magnitude of
resultant intensity at point $P$ is given by →
$ E=E_{1}+E_{2}$
$ E=\frac{\sigma_{1}}{2 \epsilon_{0}}+\frac{\sigma_{2}}{2
\epsilon_{0}}$
$ E=\frac{1}{2 \epsilon_{0}} \left (\sigma_{1}+\sigma_{2} \right )$
If both sheets have equal charge densities $\sigma$ i.e.
$\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:
$ E=\frac{\sigma}{ \epsilon_{0}} $
The electric field intensities at point $P''$ →
$ E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$ (away from sheet 1)
$ E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$ (toward from sheet 2)
The magnitude of the resultant electric field $E$ →
$ E=E_{1}-E_{2}$
$E= \frac{1}{2\epsilon_{0}} \left ( \sigma_{1}- \sigma_{2}\right )$
If both sheets have equal charge densities $\sigma$ i.e.
$\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:
$ E=0 $
From the above expression, we can conclude that the magnitude of $E$ is free
from the 'position' of the point taken in the electric field between the
sheet and outside the sheet. It is also shown that the electric field
between the sheet is uniform everywhere and independent of separation
between the sheets.
