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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

Sagot :

Given:

The force between two charged balls is,

[tex]F=6.0\times10^{-6}\text{ N}[/tex]

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

To find:

The new force between the charged balls

Explanation:

Let, the charges are,

[tex]q_1\text{ and q}_2[/tex]

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

[tex]\begin{gathered} F=k\frac{q_1q_2}{d^2}=6.0\times10^{-6}\text{ N} \\ k=Coulomb^{\prime}s\text{ constant} \end{gathered}[/tex]

Now, the distance is

[tex]2d[/tex]

and the first charge became,

[tex]2q_1[/tex]

The new force is,

[tex]\begin{gathered} F_{new}=k\frac{2q_1q_2}{(2d)^2} \\ =k\frac{q_1q_2}{2d^2} \\ =\frac{F}{2} \\ =\frac{6.0\times10^{-6}}{2} \\ =3.0\times10^{-6}\text{ N} \end{gathered}[/tex]

Hence, the new force is,

[tex]\begin{equation*} 3.0\times10^{-6}\text{ N} \end{equation*}[/tex]

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