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>> $There are four large parallel conducting$

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 :

75042

This question has multiple correct options

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 $\frac{q _{1} + q _{2} - q _{3} - q _{4}}{2 A ε _{0}}$

C

Electric field intensity at the point Q, towards right, will be $\frac{q _{1} - q _{2} + q _{3} + q _{4}}{2 ε _{0}}$

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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Updated on : 2022-09-05

Solution

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Correct options are A) and B)

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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