Question

Coulomb's Law: Two electrons are 20.0 mm apart at closest approach. What is the magnitude of the maximum electric force that they exert on each other? (e = 1.60 × 10-19 C, k = 1/4πε0 = 9.0 109 N ∙ m2/C2)

Answer 1

Answer:

F = 5.76 x 10⁻²⁵ N

Explanation:

given,

distance between electron,r = 20 mm = 0.02 m

charge of electron, q = 1.6 x 10⁻¹⁹ C

k = 9 x 10⁹ N.m²/C²

Electric force magnitude

using electric force formula = ?

$F = \dfrac{kq^2}{r^2}$

$F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.02)^2}$

     F = 5.76 x 10⁻²⁵ N

hence, the electric force exerted by the electron on reach other is equal to F = 5.76 x 10⁻²⁵ N.

Answer 2

Answer:

The maximum electric force is $5.76\times10^{-25}\ N$

Explanation:

Given that,

Distance = 20.0 mm

We need to calculate the maximum electric force that they exert on each other

Using formula of electric force

$F=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q^2}{r^2}$

Where, q = charge

r = radius

Put the value into the formula

$F=9\times10^{9}\times\dfrac{(1.6\times10^{-19})^2}{(20.0\times10^{-3})^2}$

$F=5.76\times10^{-25}\ N$

Hence, The maximum electric force is $5.76\times10^{-25}\ N$