Two equal positive point charges are separated by a distance . [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Two equal positive point charges are separated by a distance $2 a$ . The distance of a point from the centre of the line joining the two charges on the equatorial line (perpendicular bisector), at which the force experienced by a test charge $q_{0}$ becomes maximum, is $\frac{a}{\sqrt{x}} .$ The value of $x$ is __________ . [2023]

Ans.
(2)

$F = \frac{2 K q q_{0} x}{{(x^{2} + a^{2})}^{3 / 2}}$

$For F to be maximum, \frac{d F}{d x} = 0$

$\Rightarrow x = \frac{a}{\sqrt{2}}$

Want Us to Email you About Special Offers & Updates?