badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12th pass

                        

Q .1 dt 5 Oct 17: *Two point charges are separated by certain distance in air medium & experiences a force of F newton. If the charge strength of each charge is increased by 20% and distance between them reduced by 10 % ,what will be new force in terms of F ?*

3 years ago

Answers : (1)

Shailendra Kumar Sharma
188 Points
							
F=kq1q2/r2 
Now the charges are increased by 20% and distance reduced by 10% 
F*= k(1.2q1)(1.2q2)/(.9r)2 =1.6kq1q2/r2 
So F* =1.6F
 
 
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details