Question

Two particles experience an attractive electric force of 5 N. If one of the charges triples and
the distance between them doubles, what is the new electric force?

Answer 1

3.75 N

Explanation:

F = k(q)(q)/r^2 = 5 N

F' = k(q)(3q)/(2r)^2

= k×3(q)(q)/4r^2

= (3/4)[k(q)(q)/r^2]

= (3/4)F

= (3/4)(5 N)

= 3.75 N

Answer 2

The new electric force is 3.75 N.

The force on the charges by Coulomb's law is given by;

F = kq1q2/r^2

where q1=q2

F = kq1^2/r^2

Where F = 5N

If one of the charges triples;

q2 = 3q1

distance doubles = r = 2r

F' = 3kq1^2/4r^2

So;

F' = 3/4 F

F' = 3/4 × 5 = 3.75 N

The new electric force is 3.75 N.

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