# 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?

Navjyot Kalra
8 years ago
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.
Apoorva Arora IIT Roorkee
8 years ago
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
Apoorva Arora IIT Roorkee
8 years ago
Force is given by
Thanks and Regards
Apoorva Arora
IIT Roorkee