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A two point charges 4q and -q are fixed ...

A two point charges 4q and -q are fixed on the x - axis `x=-d/2 and x = d/2` , respectively. If a third point charge 'q' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will

A

Increase by `(3q^(2))/(4piepsilon_(0)d)`

B

Increase by `(2q^(2))/(3piepsilon_(0)d)`

C

Decrease by `(q^(2))/(4piepsilon_(0)d)`

D

Decrease by `(4q^(2))/(3piepsilon_(0)d)`

Text Solution

Verified by Experts

The correct Answer is:
D
Potential at O.
`impliesV_(O)=(K4q)/((d)/(2))+(K(-q))/((d)/(2))=(6Kd)/(d)`
Potential at P,
`impliesV_(P)=(K4q)/((3d)/(2))+(K(-q))/((d)/(2))=(2kq)/(3d)`
Change in potential energy of a charge (q)
`=qDeltaV=q(V_(f)-V_(i))`
`=q(V_(P)-V_(O))`
`q((2Kq)/(3d)-(6Kq)/(d))=-(16q^(2))/(4piepsilon_(0)3d)=-(4q^(2))/(3piepsilon_(0)d)`.
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