Question

1. two charges that are both - 3.0 c push each other apart from 2 meters away. what is the force? 10. a negative charge of - 7.0 x 10^{-11} c and a positive charge of 1.7 x 10^{-20} c are separated by 0.0015 m. what is the force between the two charges?

Explanation:

Step1: <Use Coulomb's Law>

Coulomb's Law is $F = k\frac{q_1q_2}{r^2}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them.

Step2: <Substitute values for problem 9>

For problem 9, $q_1=- 3.0\ C$, $q_2=-3.0\ C$, $r = 2\ m$.
$F=9\times10^{9}\times\frac{(-3.0)\times(-3.0)}{2^{2}}$
$F = 9\times10^{9}\times\frac{9}{4}$
$F=2.025\times10^{10}\ N$

Step3: <Substitute values for problem 10>

For problem 10, $q_1=-7.0\times10^{-11}\ C$, $q_2 = 1.7\times10^{-20}\ C$, $r=0.0015\ m$.
$F=9\times10^{9}\times\frac{(-7.0\times10^{-11})\times(1.7\times10^{-20})}{(0.0015)^{2}}$
First calculate the numerator: $(-7.0\times10^{-11})\times(1.7\times10^{-20})=-11.9\times10^{-31}$
The denominator: $(0.0015)^{2}=2.25\times10^{-6}$
$F = 9\times10^{9}\times\frac{-11.9\times10^{-31}}{2.25\times10^{-6}}$
$F=\frac{9\times(- 11.9)\times10^{9 - 31+6}}{2.25}$
$F=\frac{-107.1\times10^{-16}}{2.25}$
$F=-4.76\times10^{-15}\ N$

Answer:

For problem 9: $2.025\times10^{10}\ N$
For problem 10: $-4.76\times10^{-15}\ N$