What would the force be if the separation between the two charges in the top window was adjusted to 8.19 ✕10-11 m? (The animatio

Question

3 years ago

6

n will not adjust that far--you will have to calculate the answer).
q1 = q2 = 1.00 ✕ e

1 answer:

dalvyx [7] · Answer 1 · 2022-07-27T02:54:53+03:00

The electrostatic force between the two charges is $3.4\cdot 10^{-8}N$

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have the following data:

$q_1 = q_2 = 1.00e$ is the magnitude of the two charges, where

$e=1.6\cdot 10^{-19}C$ is the fundamental charge

$r=8.19\cdot 10^{-11}m$ is the separation between the two charges

Substutiting into the equation, we find the force:

$F=(8.99\cdot 10^9)\frac{(1.00\cdot 1.6\cdot 10^{-19})^2}{(8.19\cdot 10^{-11})^2}=3.4\cdot 10^{-8}N$

Learn more about electric force:

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