Two point charges one 4 times as strong as the other are placed at a certain distance apart in air. The force between them is 0.4 N. When the distance between them is decreased by 10 cm, the force is increased by 0.5 N. Calculate the charges and their original separation.

q1 = qq2 = 4qoriginal separation is r.F = kq1q2r20.4 = 4q2r20.5= 4q2(r-10)245=r-10r20.89 = r-10r0.89r=r-1010 = 0.11rr = 90.09 cmq = 0.28 C

  • -4
charge is 10raise to power -12 and distance between two charge is 30cm
  • 2
10raise to power minus 6
  • 6
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