A charge of 1.0 C is oplaced at the top of 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 teimes of your weight is this force?

Open in App

Solution

$q_{1} = q_{2} = q = 1. O C$
Distance between the charges
$= 2 k m = 2 \times 10^{3}$
So, force $= k . \frac{q_{1} q_{2}}{r^{2}}$
$F = (9 \times 10^{9}) \times \frac{1.1}{(2 \times 10^{3})^{2}} 2.25 \times 10^{3} N$
Teh weight of the body
$= m g = 40 \times 10 N = 400 N$
$= \frac{(weight of body)}{(force between charges)}$
$= \frac{1}{5.6}$
So, force between charge
= 5.6 times weight of body

Suggest Corrections

Similar questions

At what separation should two equal charges, 1.0 C each be placed so taht the force between them equals the weight of a 50 kg person?

At what distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight?

F

2

2 F

2

Join BYJU'S Learning Program