Question 2: A charge of 1.0 C is placed 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 times of your weight is this force?
Answers
Answered by
0
Answer:
The force of attraction between two charges separated by distance r is given by: F = kq1q2r2 (where k = 14πϵ0 = 9 × 109Nm2/C2 ) F = 9 × 109 × 1 ×
Answered by
25
SOLUTION :-
=>q1 = q2 = 1.0 C (Given)
=>Let “r” be the distance between the charges.
=> r = 2 km = 2 x 10³ m
By Coulomb’s law,
Using equation (A)
=>F =9 x 10^9x (1×1)/(2×10³)² =2.25x10³N
=>Again, we know W = mg
=>Let the mass of my body, m = 70 kg.
=>W = 70 x 9.8 = 686 N
=>Now, on dividing the Electrostatic force and the body weight, we have
=>F/W = [2.25 x 10³]/686 = 3.3 (approx).
The electric force between the two charges is 3.3 times my weight.
Similar questions