Question

Two charged objects attract each other with a force 1.0 N. What happens to the force between them if one charge is increased by a factor of 2, the other charge is increased by a factor of 4, and the separation distance between their centers is reduced to 1/4 its original value

Answer 1

Answer:

F' = 128 N

Explanation:

The electrostatic force of attraction between two charges is given by Colomb's Law, as follows:

$F = \frac{kq_1q_2}{r^2}$

where,

F = Force of attraction = 1 N

G = universal gravitational constant

q₁ = magnitude of the first charge

q₂ = magnitude of the second charge

r = distance between charges

Therefore,

$1\ N = \frac{kq_1q_2}{r^2}$ --------------------- eq(1)

Now, we apply the changes given in the question:

$F' = \frac{k(2q_1)(4q_2)}{(\frac{1}{4}r)^2}\\\\F' = 128(\frac{kq_1q_2}{r^2})$

using eq (1):

F' = 128(1 N)

F' = 128 N