Two point charges 4µC and 9µC are separated by 50 cm. The potential at the point between them where the field has zero strength is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

9 × 10⁵V

4.5 × 10⁵V

9 × 10⁴V

Zero

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

$x = \frac{50}{\sqrt{\frac{9}{4}} + 1} = 20 cm$ from 4µC At that point
$\begin{array}{l} V = 9 \times 10^{9} [\frac{4 \times 10^{- 6}}{20 \times 10^{- 2}} + \frac{9 \times 10^{- 6}}{30 \times 10^{- 2}}] \\ V = 4 .5 \times 10^{5} volt \end{array}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th