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

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

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

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!