Question

Two point charges 5×10^-19 and 20×10^-19 seperated by a distance 2m .Calculate the force

Answer 1

Answer:

Given,

q1 = $5*10^{-19}C$

q2 =  $20*10^{-19}C$

Distance between the two charges = r = 2 m

As we know,

F = $\frac{1}{4\pi\epsilon_0}\frac{q1q2}{r^{2}}$

    = $9*10^{9}*\frac{5*10^{-19}*20*10^{-19}}{2^{2} }$

   = $45*10^{-29}$

 

    = $4.5*10^{-28} N$

Thus the force between the two point charges is $4.5*10^{-28} N$

Additional information :

According to Coulomb's law ,

Two stationary point charges repel or attract each other with a  force which is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force is along the line joining the two charges.

Thus , If there is two point charges  q1 and q2 situated at a distance r apart , then the force acting between them is

F ∝ $\frac{q1q2}{r^{2} }$

F = k$\frac{q1q2}{r^{2} }$

here ,

k = proportionality constant = $\frac{1}{4\pi\epsilon_0}$ = $9*10^{9}$