Question

Two positive point charges, each with charge q, separated by a distance d, repel each other with a force of magnitude 20 N. What is the magnitude of the force between two positive point charges of magnitude 11.11 q, separated by a distance 2.5 d in units of N

Answer 1

The value is
`F_2 = 395 \ N`

Step-by-step explanation:

From the question we are told that

The magnitude of the charge of each positive charge for the first case is
`q_1 = q_2 = q`

The distance between the charges for the first case is
`d`

The force between the charges for the first case is
`F = 20 \ N`

The magnitude of the charge of each positive charge for the second case is
`q_1 = q_2=11.11q`

The distance between the charge for the second case is
`2.5d`

Generally for the first case the force between the charge is mathematically represented as

`F_1 = (k * q^2 )/(d^2)`

Where k is the Coulomb constant with value
`k = 9*10^(9) \ kg\cdot m^3\cdot s^(-4) \cdot A^(-2).`

So

`20= (k * q^2 )/(d^2)`

Generally for the second case the force between the charge is mathematically represented as

`F_2 = (k * (11.11q)^2)/((2.5d)^2)`

`F_2 = (k * 11.11^2 *q^2)/(2.5^2d^2)`

=>
`F_2 = (11.11^2)/(2.5^2) F_1`

=>
`F_2 = (11.11^2)/(2.5^2) * 20`

=>
`F_2 = 395 \ N`