Two particles, each having a mass of 5g and charge 10^–7 C, stay in limiting equilibrium on a horizontal table with a separation of 10cm between them. Find the coefficient of friction between each particle and the table, which is same between each particle and table

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$μ = 0 .16$

$μ = 0 .20$

$μ = 0 .18$

$μ = 0 .22$

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

Repulsion due to their charge = friction due to table.

$∴ \frac{{kQ}^{2}}{r^{2}} = μmg; Q = 10^{- 7} C r = 0.1 m \Rightarrow \frac{(9 \times 10^{9}) {(10^{- 7})}^{2}}{(0.1)^{2}} = μ \times (5 \times 10^{- 3}) (10) ∴ μ = 0.18$

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

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!