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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 q1,q2,q3 and q4 respectively :
75042_dc90adfa3b7c421db3630564b31ace13.png
  1. Charge appearing on the left hand side of plate 1 will be q1+q2+q3+q42
  2. Electric field intensity at the point P, towards right, will be q1+q2q3q42Aε0
  3. Electric field intensity at the point Q, towards right, will be q1q2+q3+q42ε0
  4. Electric field at P will be equal to that at Q, for any values of q1,q2,q3 and q4

A
Electric field intensity at the point Q, towards right, will be q1q2+q3+q42ε0
B
Electric field at P will be equal to that at Q, for any values of q1,q2,q3 and q4
C
Charge appearing on the left hand side of plate 1 will be q1+q2+q3+q42
D
Electric field intensity at the point P, towards right, will be q1+q2q3q42Aε0
Solution
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Consider a point inside the q1 plate. The electric field at that point is 0.
x(q1x)(xq1)(q2+q1x)(xq1q2) (q3+q2+q1x)(xq1q2q3)(q1+q2+q3+q4x)=0
x=(q1+q2+q3+q4)2
Also, electric field inside a field is only due to the charges on the inside of the plate.
EP=1ϵ0×(q1+q2q3q4)2
EQ=1ϵ0×(q1+q2+q3q4)2
93955_75042_ans.png

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