Question

what is the electric force acting between two charges of 0.0072 c and -0.0060 c that are 0.0040 m apart? use ( f_{e}=\frac{k q_{1} q_{2}}{r^{2}} ) and ( k = 9.00 \times 10^{9} mathrm{~n} cdot mathrm{m}^{2} / mathrm{c}^{2} ). a. ( -2.4 \times 10^{10} mathrm{~n} ) b. ( -9.7 \times 10^{7} mathrm{~n} ) c. ( 2.4 \times 10^{10} mathrm{~n} ) d. ( 9.7 \times 10^{7} mathrm{~n} )

Explanation:

Step1: Substitute the values into the formula

Given \(q_1 = 0.0072\space C\), \(q_2=- 0.0060\space C\), \(r = 0.0040\space m\), \(k = 9.00\times10^{9}\space N\cdot m^{2}/C^{2}\)

Substitute into \(F_{e}=\frac{kq_{1}q_{2}}{r^{2}}\)

\(F_{e}=\frac{(9.00\times 10^{9})\times(0.0072)\times(- 0.0060)}{(0.0040)^{2}}\)

Step2: Calculate the numerator and denominator

First, calculate the numerator: \((9.00\times 10^{9})\times(0.0072)\times(- 0.0060)=9\times10^{9}\times(- 4.32\times10^{-5})=-3.888\times10^{5}\)

Then, calculate the denominator: \((0.0040)^{2}=1.6\times10^{-5}\)

Step3: Calculate the value of \(F_{e}\)

\(F_{e}=\frac{-3.888\times 10^{5}}{1.6\times10^{-5}}=-2.43\times10^{10}\approx - 2.4\times10^{10}\space N\)

Answer:

A. \(-2.4\times 10^{10}\space N\)