Redistribution of Charges for Two Parallel Plates
Let’s talk about the redistribution of charges.
When two charged bodies are placed near each other, charges induced on them due to one another.
Due to the charges induced, the configuration of the charges will change on either body.
This phenomenon is called the redistribution of the charges.
During this phenomenon, some energy is lost in the form of heat.
To analyse this wastage of energy and new charge on bodies, we study charge redistribution.
Let’s see the redistribution of the charges in parallel plates.
Consider two plates,
$I$
&
$II$
, placed parallel to each other. Let the area of each plate is
$A$
.
The plates
$I$
and
$II$
are carrying charges
$+q_{1}$
and
$+q_{2}$
of different magnitude.
Let’s the
$q_{a}$
and
$q_{b}$
be the charges after redistribution on the left and right surface of the plate
$I$
.
And
$q_{c}$
and
$q_{d}$
be the charges after redistribution on the left and right surface of the plate
$II$
.
For plate
$I$
the total charger
$q_{1}$
will be the sum of
$q_{a}$
and
$q_{b}$
.
And for plate
$II$
the total charge
$q_{2}$
will be the sum of
$q_{c}$
and
$q_{d}$
.
Now consider a point
$P$
inside the plate
$I$
.
As the electric field inside the conductor is always zero, electric field at P
$(E_{p})$
will be zero.
Let,
$E_{1}$
,
$E_{2}$
,
$E_{3}$
and
$E_{4}$
be the electric fields due to the charges
$q_{a}$
,
$q_{b}$
,
$q_{c}$
and
$q_{d}$
respectively.
So, the electric fields at point P can be represented as
So, the total electric field at point
$P$
will be given by,
The electric field due to the charged sheet can be expressed as given above.
So, the electric field at point P can be expressed as,
Replace
$q_{b}$
and
$q_{d}$
, in the equation written for the electric field at the point
$P$
inside the first plate.
After substitution we got,
On simplifying, we will get the charge on the left surface of the first plate.
Therefore, the charge on the right surface of the first plate will be,
Now take a point Q inside the plate
$II$
.
Similar to the point P, we can represent the directions of the electric field at Q, due to charges
$q_{1},q_{2},q_{3}$
&
$q_{4}$
.
Thus, the total electric field at point
$Q$
will be,
On simplifying the equation, we get
On substituting the value of
$q_{b}$
and
$q_{d}$
in the expression, we get
On solving the equation we will get the expression of charge on the left surface of the plate
$II$
.
Thus, the charge on the right surface of the second plate will be,
Revision
The charge on the outer surfaces of both the plates are equal, and equal to the mean of the charge supplied to the plates.
The charge on the outer surfaces of the plates is,
The charge on the surfaces facing to each other is equal and opposite, and is equal to the,
The charge of the capacitor is equal to the charge on the inner surface on the positive plate.
If both the plates are considered as capacitor, the charge
$q_{b}$
will be the charge of the capacitor.
The End