Two point charges of 20 $μ$ C and 80 $μ$ C are 10 cm apart. Where will the electric field strength be zero on the line joining the charges from 20 $μ$ C charge?

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

0.33 m

0.1 m

0.033 m

0.04 m

answer is C.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let neutral point be obtained at a distance x from 20 $μ$ C charge. Hence, at neutral point

$\frac{1}{4 π ε_{0}} \frac{20 \times 10^{- 6}}{x^{2}} = \frac{1}{4 π ε_{0}} \frac{80 \times 10^{- 6}}{(0.10 - x)^{2}}$

$\frac{20}{x^{2}} = \frac{80}{(0.10 - x)^{2}}$

$x = + 0.033 m$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test