The force between two point charges is 10N. If their charge is doubled and the distance between them is reduced to half, what will be the magnitude of the force between them?

The force between two point charges is given by Coulomb’s Law:

F=k.q1q2/r2

where:

  • F is the force between the charges,
  • k is Coulomb’s constant,
  • q1 and q2 are the magnitudes of the charges, and
  • r is the distance between them.

Now, in this case, both charges are doubled (q1′=2q1 and q2′=2q2), and the distance between them is halved (r′=r/2).

The new force F will be:

F′=k(2q1)(2q2)/(r/2)2

Simplifying this expression:

F′=k.4q1q2/r2/4

=4k.q1q2/r2×4=16×(k.q1q2/r2)

Since the original force is F=kq1q2/r2=10 N, we substitute that into the equation:

F′=16×10=160 N

Therefore, the new force between the charges will be 160 N.

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