Question

what is the electric force acting between two charges of -0.0085 c and -0.0025 c that are 0.0020 m apart? use ( f_{e}=\frac{k q_{1} q_{2}}{r^{2}} ) and ( k = 9.00\times10^{9}ncdot m^{2}/c^{2} ). a. ( 4.8\times10^{10}n ) b. ( 9.6\times10^{7}n ) c. ( -9.6\times10^{7}n ) d. ( -4.8\times10^{10}n )

Explanation:

Step1: Substitute the values into the formula

Given \(k = 9.00\times10^{9}\space N\cdot m^{2}/C^{2}\), \(q_{1}=- 0.0085\space C\), \(q_{2}=-0.0025\space C\), \(r = 0.0020\space m\)

$$ F_{e}=\frac{kq_{1}q_{2}}{r^{2}}=\frac{(9.00\times 10^{9})(-0.0085)(-0.0025)}{(0.0020)^{2}} $$

Step2: Calculate the numerator

\((9.00\times 10^{9})(-0.0085)(-0.0025)=9.00\times10^{9}\times0.0085\times0.0025 = 9.00\times10^{9}\times2.125\times10^{- 5}=1.9125\times10^{5}\)

Step3: Calculate the denominator

\((0.0020)^{2}=4\times10^{-6}\)

Step4: Calculate the force

$$ F_{e}=\frac{1.9125\times 10^{5}}{4\times10^{-6}}=\frac{1.9125}{4}\times10^{5 + 6}=0.478125\times10^{11}=4.78125\times10^{10}\approx4.8\times10^{10}\space N $$

Since both charges are negative, the force is repulsive. The magnitude is considered in terms of the formula result (the negative signs cancel out when calculating the magnitude of the force as we are interested in the size of the interaction).

Answer:

A. \(4.8\times 10^{10}\space N\)