wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A certain charge Q is divided into two parts q and Qq. How the charge Q and q must be related so that when q and (Qq) is placed at a certain distance apart experience maximum electrostatic repulsion?

A
Q=2q
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Q=3q
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Q=4q
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Q=4q+c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A Q=2q
The electrostatic force of repulsion between the charge q and (Qq) at separation r is given by
F=14πε0 q(Qq)r2 =14πε0 qQq2r2
If F is maximum, then F/q=0
i.e., 14πε0(Q2q)r2=0
As 14πε0r2 is constant, therefore
Q2q=0 or Q=2q

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Coulomb's Law - Children's Version
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon