Charge Q is placed at a distance r from a uniformly charged rod having 'λ' as linear charge density. The electrostatic force between them (Length of rod = L) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{kQλL}{r(r+L)}$
The charge dQ on this small segment is -
$dq=λdx=\frac{Q}{L}dx$
$∴dF=k\frac{dq.q}{x^2}=k\frac{Qq}{L}\frac{dx}{x^2}$
$∴\int dF=\int\limits_r^{r+L}k\frac{Qq}{L}\frac{dx}{x^2}$
$=\frac{kQq}{r(r+L)}=\frac{kQλL}{r(r+L)}$