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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 :

ANSWER:

18 N

STEP-BY-STEP EXPLANATION:

Force between two charge particles is given by:

[tex]F=k\frac{q_1\cdot q_2}{d^2}[/tex]

We can make the following relationship:

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

Therefore, we can establish the following:

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

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

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