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

The distance between charges 5×1011C and 2.7×1011C is 0.2m. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is

A
0.44 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.65 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.556 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
0.350 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 0.556 m

If two opposite charges are separated by a certain distance, then for it's equalibrium a third
charge should be kept outside and near the charge which is smaller in magnitude.
Here, suppose third charge q is placed at a distance x from 2.7×1011C then for it's
equalibrium |F1|=|F2|


KQ1q(x+0.2)2=KQ2qx2 x=0.556 m

(Here K=14πϵ0 and Q1=5×1011C, Q2=2.7×1011C)


flag
Suggest Corrections
thumbs-up
22
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Faraday’s Law of Induction
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon