Question

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menus to express the new force in terms of F.
q1 is halved, q2 is doubled, but the distance between the charges remains d.
F/4
F
6F
6F/4


q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
F/4
F
6F
6F/4

q1 is doubled and q2 is tripled. The distance between the charges remains d.
F/4
F
6F
6F/4

Answer 1

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

Answer 2

Answer

F) q1 is halved, q2 is doubled, but the distance between the charges remains d.

F/4) q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

6F) q1 is doubled and q2 is tripled. The distance between the charges remains d.