Question

two particles are separated by 0.38 m and have charges of $-6.25 \times 10^{-9} mathrm{c}$ and $2.91 \times 10^{-9} mathrm{c}$. use coulombs law to predict the force between the particles if the distance is cut in half. the equation for coulombs law is $f_{e}=\frac{k q_{1} q_{2}}{r^{2}}$, and the constant, $k$, equals $9.00 \times 10^{9} mathrm{~n} cdot mathrm{m}^{2} / mathrm{c}^{2}$. a. $-1.13 \times 10^{-6} mathrm{~n}$ b. $-4.53 \times 10^{-6} mathrm{~n}$ c. $4.53 \times 10^{-6} mathrm{~n}$ d. $1.13 \times 10^{-6} mathrm{~n}$

Explanation:

Step1: Determine the new distance

The original distance \(r = 0.38\space m\). When the distance is cut in half, the new distance \(r_{new}=\frac{0.38}{2}= 0.19\space m\)

Step2: Apply Coulomb's law \(F_{e}=\frac{kq_{1}q_{2}}{r^{2}}\)

We know \(k = 9.00\times10^{9}\space N\cdot m^{2}/C^{2}\), \(q_{1}=- 6.25\times10^{-9}\space C\), \(q_{2}=2.91\times10^{-9}\space C\), and \(r_{new}=0.19\space m\)

Substitute the values into the formula:

$$
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Answer:

B. \(-4.53\times 10^{-6}\space N\)