Question

question 7 of 25 what is the electric force acting between two charges of -0.0045 c and -0.0025 c that are 0.0060 m apart? use $f_{e}=\frac{kq_{1}q_{2}}{r^{2}}$ and $k = 9.00\times10^{9}ncdot m^{2}/c^{2}$. a. -2.8×10^{9}n b. -1.7×10^{7}n c. 1.7×10^{7}n d. 2.8×10^{9}n

Explanation:

Step1: Identify values

$q_1=- 0.0045\ C$, $q_2=-0.0025\ C$, $r = 0.0060\ m$, $k = 9.00\times10^{9}\ N\cdot m^{2}/C^{2}$

Step2: Substitute values into formula

$F_e=\frac{kq_1q_2}{r^{2}}=\frac{9.00\times 10^{9}\times(-0.0045)\times(-0.0025)}{(0.0060)^{2}}$

Step3: Calculate numerator

$9.00\times 10^{9}\times(-0.0045)\times(-0.0025)=9.00\times10^{9}\times0.0045\times0.0025 = 101250$

Step4: Calculate denominator

$(0.0060)^{2}=0.0060\times0.0060 = 3.6\times10^{-5}$

Step5: Calculate force

$F_e=\frac{101250}{3.6\times 10^{-5}}=2.8125\times 10^{9}\ N$
Since the charges are both negative, the force is repulsive and positive.

Answer:

D. $2.8\times 10^{9}\ N$