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        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?
6 years ago

Navjyot Kalra
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.


6 years ago
Apoorva Arora
IIT Roorkee
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 RegardsApoorva AroraIIT RoorkeeaskIITians Faculty

6 years ago
Apoorva Arora
IIT Roorkee
181 Points


Force is given by

Thanks and Regards

Apoorva Arora

IIT Roorkee


6 years ago
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### Course Features

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