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
Amit Saxena Grade: upto college level
        A charge of 1.0 C is placed at the top of the your college building and another equal charge at the top of your house. Take the separation between the two charges to be 2.0 km. Find the force exerted by the charges on each other. How many times of your weight is this force?
3 years ago

Answers : (3)

Navjyot Kalra
askIITians Faculty
654 Points
										Sol. q abse 1 = q base 2 = q = 1.0 C distance between = 2 km = 1 × 10^3 m

so, force = kq base 1q base 2/r^2 F = (9 * 10^9) * 1 * 1/(2 * 10^3)^2 = 9 * 10^9/2^2 * 10^6 = 2,25 * 10^3 N
The weight of body = mg = 40 * 10 N = 400 N
So, wt of body/force between ch arg es = (2.25 * 10^3/4 * 10^2)^-1 = 1/5.6
So, force between charges = 5.6 weight of body.
3 years ago
Apoorva Arora
IIT Roorkee
askIITians Faculty
181 Points
										
Force is given by
F=\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{r^{2}}
=9\times 10^{9}\times 1\times 1/(2000)^{2}=2250N
Thanks and Regards
Apoorva Arora
IIT Roorkee
askIITians Faculty
3 years ago
Apoorva Arora
IIT Roorkee
askIITians Faculty
181 Points
										



Force is given by


Description: F=\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{r^{2}}


Description: =9\times 10^{9}\times 1\times 1/(2000)^{2}=2250N


Thanks and Regards


Apoorva Arora


IIT Roorkee


askIITians Faculty



3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details
  • Electricity and Magnetism
  • OFFERED PRICE: Rs. 1,696
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

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