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Electric Field Due to Disk and Infinitely Large Sheet

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Question

There are four large parallel conducting plates each of Area A, placed parallel to each other. Plates 1, 2, 3, 4 are given charges equal to $q_{1}, q_{2}, q_{3}$ and $q_{4}$ respectively :

A

Charge appearing on the left hand side of plate 1 will be

\frac{q_{1} + q_{2} + q_{3} + q_{4}}{2}

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B

Electric field intensity at the point P, towards right, will be

\frac{q_{1} + q_{2} - q_{3} - q_{4}}{2 A ε_{0}}

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C

Electric field intensity at the point Q, towards right, will be

\frac{q_{1} - q_{2} + q_{3} + q_{4}}{2 ε_{0}}

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D

Electric field at P will be equal to that at Q, for any values of

q_{1}, q_{2}, q_{3}

and

q_{4}

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Solution

The correct options are
A Charge appearing on the left hand side of plate 1 will be $\frac{q_{1} + q_{2} + q_{3} + q_{4}}{2}$
B Electric field intensity at the point P, towards right, will be $\frac{q_{1} + q_{2} - q_{3} - q_{4}}{2 A ε_{0}}$
Consider a point inside the $q_{1}$ plate. The electric field at that point is 0.
$∴ x - (q_{1} - x) - (x - q_{1}) - (q_{2} + q_{1} - x) - (x - q_{1} - q_{2}) -$ $(q_{3} + q_{2} + q_{1} - x) - (x - q_{1} - q_{2} - q_{3}) - (q_{1} + q_{2} + q_{3} + q_{4} - x) = 0$
$\Rightarrow x = \frac{(q_{1} + q_{2} + q_{3} + q_{4})}{2}$
Also, electric field inside a field is only due to the charges on the inside of the plate.
$E_{P} = \frac{1}{ϵ_{0}} \times \frac{(q_{1} + q_{2} - q_{3} - q_{4})}{2}$
$E_{Q} = \frac{1}{ϵ_{0}} \times \frac{(q_{1} + q_{2} + q_{3} - q_{4})}{2}$

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Q. Four point charges

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