Join now for JEE/NEET and also prepare for Boards Class 10 Boards Cancelled! What to do next ? Ask our Experts in a Free Counseling Session Now ! Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
An electron and a proton are lying 10 cm apart The ratio of electrostatics force and gravitational force between them will be
Dear pallavi
me =9.1 * 01-31
mp = 1.6 * 10-27
qe = -1.6 *10-19
qp =1.6 *10-19
Fg =Gmemp/(10)2
Fe =1/4∏ε qeqp/(10)2
Fe/Fg = (1/4∏ε) qeqp /Gmemp
= 9*109 *(1.6 *10-19)2/(6.67*10-11 * 9.1 * 10-31 *1.6 * 10-27 )
=.23 * 10 40
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation.
All the best. Regards,Askiitians ExpertsBadiuddin
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !