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?

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Accepted Answer

# Explanation:
## Step1: Recall Coulomb's Law
Coulomb's Law is given by $F = k\frac{\vert q_1q_2\vert}{r^{2}}$, where $k = 9\times 10^{9}\ N\cdot m^{2}/C^{2}$, $q_1=-7.0\times 10^{-11}\ C$, $q_2 = 1.7\times 10^{-20}\ C$, and $r=0.0015\ m$.
## Step2: Substitute values into the formula
Substitute the values into the formula:
$$
\begin{align*}
F&=(9\times 10^{9})\frac{\vert(- 7.0\times 10^{-11})(1.7\times 10^{-20})\vert}{(0.0015)^{2}}\
&=(9\times 10^{9})\frac{(7.0\times 10^{-11}\times1.7\times 10^{-20})}{2.25\times 10^{-6}}\
&=(9\times 10^{9})\frac{11.9\times 10^{-31}}{2.25\times 10^{-6}}\
&=(9\times 10^{9})\times5.289\times 10^{-25}\
& = 4.76\times10^{-15}\ N
\end{align*}
$$

# Answer:
The force between the two charges is $4.76\times 10^{-15}\ N$

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Question

Explanation:

Step1: Recall Coulomb's Law

Step2: Substitute values into the formula

Answer: