A square of side √2 has charges of +2×10−9C,1×10−9C,−2×10−9Cand−3×10−9C respectively at its corners. The potential at the center of the square is
−8V
+8V
−18V
+18V
A
−18V
B
−8V
C
+18V
D
+8V
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Solution
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Here the potential due to 2×10−9C and −2×10−9C gets cancelled. Hence net potential isgiven by-
v=−3q4πϵ0+q4πϵ0=−2q4πϵ0=−q2πϵ0
where q=10−9C
v=(−q2πϵ0)=+18V
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