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Question

A square of side 2 has charges of +2×109C,1×109C,2×109Cand3×109C respectively at its corners. The potential at the center of the square is


  1. 8V
  2. +8V
  3. 18V
  4. +18V

A
18V
B
8V
C
+18V
D
+8V
Solution
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Here the potential due to 2×109C and 2×109C gets cancelled. Hence net potential isgiven by-
v=3q4πϵ0+q4πϵ0=2q4πϵ0=q2πϵ0
where q=109C
v=(q2πϵ0)=+18V

1225796_1026212_ans_eb6dc2b6618e465a8aab25316d3dae2c.png

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