Two positive point charges of 10   μC and 12   μC are placed 10   cm apart in air.…

Two positive point charges of 10 μC and 12 μC are placed 10 cm apart in air. The work done to bring them 6 cm closer is
  1. 8.1 J
  2. 3.2 J
  3. 9 J
  4. 13.5 J

Solution

Work done = Increase in potential energy

=Kq1q21r2-1r1

=9×109×10×10-6×12×10-616-110×102

=7.2 J

Hence, the correct option is 8.1 J.

Asked in: AP EAMCET 2022 (04 Jul Shift 1)

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