A positive charged particle of 100mg is thrown in opposite direction to a uniform electric field of strength $10^{5} N / C$ . If the charge on the particle is $40 μ C$ and the initial velocity is 200m/s, how much distance it will travel before coming to the rest momentarily.

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

0.5m

10m

answer is D.

(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

$a = \frac{F}{m} = \frac{E q}{m} = \frac{10^{5} \times 40 \times 10^{- 6}}{100 \times 10^{- 3} \times 10^{- 3}} = \frac{40 \times 100}{100 \times 10^{- 3}} = 40 \times 10^{3}$

$V^{2} - u^{2} = 2 a s$

$0^{2} - {(200)}^{2} = 2 \times 40 \times 10^{3} \times s$

$∴ S = \frac{4 \times 10^{4}}{2 \times 40 \times 10^{3}} = \frac{1}{2} = 0.5 m$

Watch 3-min video & get full concept clarity

hear from our champions

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

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

Get Expert Academic Guidance – Connect with a Counselor Today!