Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

The electric field on the axis of a dipole of dipole moment $\;\overrightarrow{P}\;$ at a distance x from the line joining the charges and $\;(x > > d)\;$

$(a)\;\large\frac{k\overrightarrow{P}}{x^3}\qquad(b)\;-\large\frac{k\overrightarrow{P}}{x^3}\qquad(c)\;\large\frac{2 k\overrightarrow{P}}{x^3}\qquad(d)\;-\large\frac{2 k\overrightarrow{P}}{x^3}$

Can you answer this question?

1 Answer

0 votes
Answer : (b) $\;-\large\frac{k\overrightarrow{P}}{x^3}$
Explanation :
$E_{D}=\large\frac{2 k\;q }{((\large\frac{d}{2})^{2}+x^2)^{\large\frac{3}{2}}}\times \large\frac{d}{2}$
This is because only horizontal components will add vertical cancel each other being equal and opposite
$E_{D} \approx \large\frac{2 \;k \;q\; d}{2x^3}$
$E_{D} \approx \large\frac{k \; P}{x^3}$
From the diagram it is clear the $\;E_{D} \;$ & P are in opposite direction


answered Feb 7, 2014 by yamini.v
edited Aug 14, 2014 by thagee.vedartham

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App