Identical charges each of value q are placed at the four corners A, B, C and D of a square of side a. The magnitude of the force on the charge at any one corner of the square is
A
3q24πϵ0a5
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B
q24πϵ0a2
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C
(1+2√22)q24πϵ0a2
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D
(1+2√2√2)q24πϵ0a2
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Solution
The correct option is C(1+2√22)q24πϵ0a2
F12=F14=q24πϵ0a2 and F12⊥F14 Resultant of F12 and F14 is √2q24πϵ0a2 along F13 Further F13=q24πϵ0(√2a)2 So total force on 1 is F1=F12+F13 ⇒F1=q24πϵ0a2(12+√2)⇒F1=(1+2√22)q24πϵ0a2