Two identical thin rods each of length 2a carry equal charges +Q that is distributed uniformly along their lengths. The rods lie along the x –axis with their centers separated by a distance b > 2a.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Force on the right rod due to the left rod is $\Rightarrow F = \frac{Q^{2}}{16 {πε}_{0} a^{2}} \log_{e} (\frac{b^{2}}{b^{2} - 4 a^{2}})$

Force on the right rod due to the left rod is $\Rightarrow F = \frac{Q^{2}}{8 \sqrt{2} {πε}_{0} a^{2}} \log_{e} (\frac{b^{2}}{b^{2} - 4 a^{2}})$

Field due to left rod at a distance x from the right end is $\Rightarrow E = \frac{Q}{4 {πε}_{0} x (x + 2 a)}$

Field due to left rod at a distance x from the right end is $\Rightarrow E = \frac{Q}{2 {πε}_{0} x (x + 2 a)}$

answer is A, B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Field on a point at a distance of 'x' from the right end is given by

$dE = \frac{dq}{4 {πε}_{0} x^{2}} = (\frac{λ}{4 {πε}_{0}}) x^{- 2} dx$
$\Rightarrow E = \frac{λ}{4 {πε}_{0}} = (\int_{d}^{d + 2 a} x^{- 2} dx) = \frac{λ (2 a)}{4 {πε}_{0} (d + 2 a)}$ $\Rightarrow E = \frac{Q}{4 {πε}_{0} d (d + 2 a)}$

The force on the right rod is given by

$dF = Edq = \frac{Q (λdx)}{4 {πε}_{0} x (x + 2 a)}$
Since $λ = \frac{Q}{2 a}$
$\Rightarrow dF = (\frac{Q}{4 {πε}_{0}}) (\frac{Q}{2 a}) [\frac{dx}{x (x + 2 a)}]$
$\Rightarrow F = \int dF = \frac{Q^{2}}{8 {πε}_{0} a} \int_{b - 2 a}^{b} \frac{dx}{x (x + 2 a)}$
$\Rightarrow F = \frac{Q^{2}}{16 {πε}_{0} a^{2}} \log_{e} (\frac{b^{2}}{b^{2} - 4 a^{2}})$