1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Two charges 9e and 3e are placed at a separation r. The distance of the point where the electric field intensity will be zero, is-

A
r(1+3) from 9e charge
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
3r3+1 from 9e charge
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
r(13) from 3e charge
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
3r(1+3) from 3e charge
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is B √3r√3+1 from 9e chargeTwo charges 9e and 3e separation =rThe distance of the point where the electric field intensity will be =0Let, distance at which electric field intensity =0 be ′a′ from 9e charge. So, Putting formula, the electric field for 3e and 9e will be equal.14πϵ09e×1a2=14πϵ0×3e(r−a)2⇒3(r−a)2=a2Solving we get,√3(r−a)=aor, √3r−√3a=aor, √3r=a(1+√3)or, a=r√31+√3from 9e charge.

Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
The Electric Field
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program