Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Two identical conducting spheres R and S have negative charges $Q_{1} and Q_{2}$ respectively, but $Q_{1} \neq Q_{2}$ . The spheres are brought to touch each other and then kept in their original positions, now the force between them 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

greater than that before the spheres touched

less than that before the spheres touched

same as that before the spheres touched

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

$F_{final} = \frac{{KQ}_{1} Q_{2}}{R^{2}}$

After contact charge on both sphere will be $- (\frac{Q_{1} + Q_{2}}{2})$

Final

$F_{final} = \frac{K {\{\frac{Q_{1} + Q_{2}}{2}\}}^{2}}{R^{2}}$

As we know for two numbers

Arithmetic mean > Geometric mean

So, $\frac{Q_{1} + Q_{2}}{2} > \sqrt{Q_{1} Q_{2}}$

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!