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: upto college level
        At what separation should two equal charge, 1.0 C each, be placed so that the force between them equals the weight of a 50 kg person?
4 years ago

Answers : (3)

Deepak Patra
askIITians Faculty
471 Points
							Sol. q = 1 C, Let the distance be x
F = 50 * 9.8 = 490
F = kq^2/x^2 ⇒ 460 = 9 * 10^9 * 1^2/x^2 or x^2 = 9 * 10^9/490 = 18.36 * 10^6
⇒ x = 4.29 * 10^3 m

						
4 years ago
Apoorva Arora
IIT Roorkee
askIITians Faculty
181 Points
							
a 50 kg person exerts the force, F=mg=500 newtons assuming g=10 m/s(square).
So,
F=500= 9\times 10^{9}\times 1\times 1/r^{2}
r^{2}=18\times 10^{6}
so r\approx 4.24\times 10^{3}metres
nearly 4.24 kilometres
Thanks and Regards
Apoorva Arora
IIT Roorkee
askIITians Faculty
4 years ago
Apoorva Arora
IIT Roorkee
askIITians Faculty
181 Points
							
a 50 kg person exerts the force, F=mg=500 newtons assuming g=10 m/s(square).
So,
Description: F=500= 9\times 10^{9}\times 1\times 1/r^{2}
Description: r^{2}=18\times 10^{6}
so Description: r\approx 4.24\times 10^{3}metres
nearly 4.24 kilometres
Thanks and Regards
Apoorva Arora
IIT Roorkee
askIITians Faculty
4 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

  • 57 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