Question

Two charges separated a distance of 1.0 meter exert a 2.0-N force on each other. If the charges are pushed to a separation of 1/3 meter, the force on each charge will be?

Answer 1

18 N

Explanation:

Force between two charge particles is given by:

`F=k(q_1\cdot q_2)/(d^2)`

We can make the following relationship:

`\begin{gathered} F_1=k(q_1\cdot q_2)/(d_1^2) \\ \\ F_2=k(q_1\cdot q_2)/(d_2^2) \\ \\ (F_2)/(F_1)=(k(q_1\cdot q_2)/(d_2^2))/(k(q_1\cdot q_2)/(d_1^2)) \\ \\ (F_2)/(F_1)=(d_2^2)/(d_1^2) \\ \\ (F_(2))/(F_(1))=\left((d_1)/(d_2)\right)^2 \end{gathered}`

Therefore, we can establish the following:

`\begin{gathered} (F_2)/(F_1)=\left((d_1)/(d_2)\right)^2 \\ \\ \text{ We replacing:} \\ \\ (F_2)/(2)=\left((1)/((1)/(3))\right)^2 \\ \\ F_2=2\cdot9 \\ \\ F_2=18\text{ N} \end{gathered}`

Therefore, the force will now be 18 N at each charge.