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Electromagnetic Induction> A charge of 1.0 C is placed at the top of...

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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?

Amit Saxena , 11 Years ago

Grade upto college level

3 Answers

Navjyot Kalra

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.

Last Activity: 11 Years ago

Apoorva Arora

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

Last Activity: 11 Years ago

Apoorva Arora

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

Last Activity: 11 Years ago

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