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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
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
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

3 years ago
Apoorva Arora
IIT Roorkee
181 Points
										 Force is given byThanks and RegardsApoorva AroraIIT RoorkeeaskIITians Faculty

3 years ago
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