Question

3.1.4 quiz: noncontact forces
question 8 of 10
two particles are separated by 0.38 m and have charges of -6.25×10^(-9) c and 2.91×10^(-9) c. use coulombs law to predict the force between the particles if the distance is doubled. the equation for coulombs law is (f_{e}=\frac{kq_{1}q_{2}}{r^{2}}), and the constant, k, equals 9.00×10^9 n·m²/c².
a. -2.83×10^(-7) n
b. -1.13×10^(-6) n
c. 2.83×10^(-7) n
d. 1.13×10^(-6) n

Explanation:

Step1: Identify given values

$k = 9.00\times 10^{9}\ N\cdot m^{2}/C^{2}$, $q_1=-6.25\times 10^{-9}\ C$, $q_2 = 2.91\times 10^{-9}\ C$, new $r = 2\times0.38\ m= 0.76\ m$

Step2: Apply Coulomb's law

$F_e=\frac{kq_1q_2}{r^{2}}=\frac{9.00\times 10^{9}\times(- 6.25\times 10^{-9})\times(2.91\times 10^{-9})}{(0.76)^{2}}$
$=\frac{9.00\times(- 6.25)\times(2.91)\times10^{9 - 9-9}}{0.5776}$
$=\frac{-163.6125\times10^{-9}}{0.5776}\approx - 2.83\times 10^{-7}\ N$

Answer:

A. $-2.83\times 10^{-7}\ N$