AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

The figure below represents the three points charges forming equilateral triangle.
The energy system of three charges,
U=3(‌

kq1q2

r

)
U=3×‌

k(2×10−3×2×10−3)

1∕2

U=24k×10−6J
When one charge move to the midpoint of the line joining the other two charges,
The energy of the system,
U′‌‌=‌

k(2×2×10−6)

(1∕4)

×2+‌

k(2×2×10−6)

(1∕2)

U′‌‌=40k×10−6T
The net energy,
∆U=U′−U
∆U=(40−24)k×10−6
∆U=16×9×103J
The energy supplied to the system at the rate of 2kW