Answered

What is the magnitude of the electrostatic force

between two electrons separated by a distance of

1.00 × 10^–8 meter?

(1) 2.56 × 10^–22 N (3) 2.30 × 10^–12 N

(2) 2.30 × 10^–20 N (4) 1.44 × 10^–1 N

Verified Answer :

The electron charge is called the Elementary Charge, given by:

\(e=1.6*10^{-19}\)
 
     The Eletrostatic Force bewteen two changes is giver by Coulomb Law:

\(F= \frac{Qqk}{d^2}\)
 
     Substituing the unknowns, we have:

\(F= \frac{(1.6*10^{-19})^2}{(10^{-8})^2} \\ F=(1.6*10^{-11})^2 \\ \boxed {F=2.56*10^{-22}N}\)

If you notice any mistake in my english, please let me know, because i am not native.

The correct option is option (1)

The magnitude of the electrostatic force is 2.56 × 10⁻²² N

Electrostatic force:

The electrostatic force on two identical charges q separated by a distance r is given by:

\(F = \frac{kq^2}{r^2}\)

where k is Coulomb's constant

q is the charge, and

r is the separation between the charges.

For an electron:

q = 1.6 × 10⁻¹⁹ C

and it is given that the two electrons are separated by a distance:

r = 1 × 10⁻⁸ m

So the electrostatic force between them will be:

\(F=\frac{kq^2}{r^2}=\frac{9\times10^9\times(1.6\times10^{-19})^2}{(1\times10^{-8})^2}\\\\F=2.56\times10^{-22}\;N\)

Learn more about electrostatic force:

https://brainly.com/question/9774180?referrer=searchResults

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