KVPY SA Exam 2014 Question Paper

Question : 25 of 80

Marks: +1, -0

Two equal charges of magnitude O each are placed at a distance d apart. Their electrostatic energy is E. A third charge

-Q/2

is brought midway between these two charges.
The electrostatic energy of the system is now

Solution:

Energy E =

{kQ× Q}/d={kQ^2}/d

….(1)
Third charge is put between them

Energy of system =

{kQ× Q}/d+{kQ}/{d/2}({-Q}/2)+{kQ}/{d/2}({-Q/2)

={kQ^2}/d+({-kQ^2}/d)+{{-kQ^2}/d}

-{kQ^2}/d

From (1)
Energy of system = – E

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