Question

When two unequal point charges are released a distance d from one another, the heavier one has an acceleration a. If you want to reduce this acceleration to one-fifth of this value, how far (in terms of d) should the charges be when released?

Answer 1

F = k(q1q2)/r^2.

To reduce the force to 1/5 by changing distance, increase the distance by sqrt(5).

So the charges should be d*sqrt(5) apart.

Answer 2

so it should be 2.236 that is the sqrt(5)
or do i have to solve it by using that equation.. because i don't have q1 and q2

Answer 3

The distance in terms of d is d*sqrt(5). Whether you have to convert it to 2.236*d depends on your teacher.

With the information given, you are not expected to solve for an actual value. Just leave it in terms of d.

Answer 4

ok thank you