Question

(a) What is the electric potential energy of two electrons separated by 2.13 nm? (b) What would be it be if the separation was doubled?

Answer 1

a) when two electrons Q₁ & Q₂ are separated by distance (r₁₂) then Electrical potential energy (U₁₂)

U₁₂= (K*Q₁*Q₂)/r₁₂ ------ Formulae

Where ,

K = Columb's constant =1/(4πo) = 9×10⁹N-m²/C²

o= 8.85×10^-12 C²/N-m²

Q₁ = Q₂ = charge of an electron = 1.602×10^-19C

r₁₂ = distance between electrons 1&2

r_{12​​​​​​}= 2.13 nm = 2.13×10^-9 m

Substitute all values in above Formulae,

We get ;

U₁₂ = {(9×10⁹)*(1.602×10^-19)²} / {2.13×10^-9)

U₁₂ = { 9×10⁹ * 2.566×10^-38} / {2.13×10^-9)

U₁₂ = 2.309×10^-28 / 2.13×10^-9

U₁₂ = 1.084×10^-19 Joules

b) As we know that potential (U₁₂) inversely proportional to distance (r₁₂)

if separation was doubled (ie.,2 times old distance ) then New potential energy (U₁₂) decrease by 2 times

U₁₂​​​​​_(new)= U₁₂/2

U_{12 (new)} = 1.084×10^-19/2

U_12(new) = 0.542×10^-19 Joules