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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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