Question

calcule the electrostatic force between two electrons that are 0.1nm apart

Answer 1

by this calculation u can find ......

Answer 2

Answer: The correct answer is $F= 23.04\times10^{-38} N$.

Explanation:

The expression for electrostatic force between the charges is as follows;

$F=(\frac{1}{4\pi \epsilon _{0}})(\frac{q_{1}q_{2}}{r^{2}})$

Here, $q_{1},q_{2}$ are the charges, $(\frac{1}{4\pi \epsilon _{0}})=9\times10^{9}\frac{Nm^{2}}{C^{2}}$ and r is the distance between the charges.

It is given in the problem that two electrons are 0.1 nm distance apart from each other.

Convert distance from nm to m.

$1\ nm= 10^{9} m$

Put $q_{1},q_{2}=1.6\times10^{-19}\ C$

$(\frac{1}{4\pi \epsilon _{0}}=9\times10^{9}\frac{Nm^{2}}{C^{2}})$

And $1\ nm= 10^{9} m$

$F=(9\times10^{9})\frac{(1.6\times10^{-19})^2}{10^9}$

$F= 23.04\times10^{-38} N$

Therefore, the electrostatic force between the given charges is $F= 23.04\times10^{-38} N$.