A 20 $μ C$ charge is divided into two parts and placed at 2 cm distance so that the repulsive force between them is maximum. The electric potential at a point where electric field strength is zero due to the parts of the charges 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

Zero

18 kV

36 kV

9 kV

answer is B.

(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

For $(I_{\max}) q_{1} = q_{2} = \frac{20}{2} μ C$
$\begin{array}{l} V = 2 (\frac{1}{4 π ε_{0}} \frac{q}{r}) \Rightarrow 2 \times 9 \times 10^{9} \times \frac{10^{- 8}}{10^{- 2}} V \\ = 18 k V \end{array}$

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!