Question

4. The force between two charged balls is 6.0 × 10–6 N. If the distance is doubled and the charge on one ball is doubled, what is the new force between the two charged balls? a. 3.0 × 10–6 N b. 6.0 × 10–6 N c. 3.0 × 10–3 N d. 6.0 × 10–3 N

Answer 1

Given:

The force between two charged balls is,

`F=6.0*10^(-6)\text{ N}`

The distance between the balls is doubled and the charge on one ball is doubled.

To find:

The new force between the charged balls

Step-by-step explanation:

Let, the charges are,

`q_1\text{ and q}_2`

The distance between the charges be d, then the force between the charged balls is,

`\begin{gathered} F=k(q_1q_2)/(d^2)=6.0*10^(-6)\text{ N} \\ k=Coulomb^(\prime)s\text{ constant} \end{gathered}`

Now, the distance is

`2d`

and the first charge became,

`2q_1`

The new force is,

`\begin{gathered} F_(new)=k(2q_1q_2)/((2d)^2) \\ =k(q_1q_2)/(2d^2) \\ =(F)/(2) \\ =(6.0*10^(-6))/(2) \\ =3.0*10^(-6)\text{ N} \end{gathered}`

Hence, the new force is,

`\begin{equation*} 3.0*10^(-6)\text{ N} \end{equation*}`