Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Four point charges $-q,+q \:,\:-q\ and \ +q$ are place on y-axis at $y=-2d,y=-d,y=+d\:\:and\:\:y=+2d$ respectively . The magnitude of the electric field $E$ at a point on the x-axis at $x=D,$ with $D\gg d$ , will behave as :

Option 1)
$E\propto \frac{1}{D^{3}}$

Option 2)
$E\propto \frac{1}{D}$

Option 3)
$E\propto \frac{1}{D^{4}}$

Option 4)
$E\propto \frac{1}{D^{2}}$

Answers (1)

best_answer

$E=\frac{2kqD}{(D^{2}+d^{2})^{3/2}}-\frac{2kqD}{(D^{2}+4d^{2})^{3/2}}$

$\Rightarrow E= 2kq D \left [ (D^{2}+d^{2})^{-3/2} -(D^{2}+4d^{2})^{-3/2}\right ]$

$\Rightarrow E= \frac{2kq D}{D^{3}} \left [ \left ( 1+\frac{d^{2}}{D^{2}} \right )^{-3/2}-\left ( 1+\frac{4d^{2}}{D^{2}} \right )^{-3/2}\right ]$

$\Rightarrow E= \frac{2kq }{D^{2}} \left [ \left ( 1+\frac{3}{2}\frac{d^{2}}{D^{2}} \right )-\left ( 1-\frac{3}{2}\frac{4d^{2}}{D^{2}} \right )\right ]$

Applying binomial approximation $\because d < < < D$

$E=\frac{9kq\:d^{2}}{D^{4}}$

$\Rightarrow E \propto \frac{1}{D^{4}}$

Option 1)

$E\propto \frac{1}{D^{3}}$

Option 2)

$E\propto \frac{1}{D}$

Option 3)

$E\propto \frac{1}{D^{4}}$

Option 4)

$E\propto \frac{1}{D^{2}}$

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions

anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question