Question

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.66 N. Determine the distance between the charges.ns

Answer 1

it is separated by 80 cm distance

Step-by-step explanation:

As per Coulombs law we know that force between two point charges is given by

`F = (kq_1q_2)/(r^2)`

here we know that

`q_1 = +8.4\mu C`

`q_2 = +5.6 \mu C`

force between two charges is given as

`F = 0.66 N`

now we have

`F = (kq_1q_2)/(r^2)`

`0.66 = ((9* 10^9)(8.4 \mu C)(5.6 \mu C))/(r^2)`

`r = 0.8 m`

so it is separated by 80 cm distance

Answer 2

d=0.8 m : Distance between the charges

Step-by-step explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

F=K*q₁*q₂/d² Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)

d: distance between the charges in meters(m)

Equivalence

1uC= 10⁻⁶C

Data

F=0.66 N

K=8.99x10⁹N*m²/C²

q₁ = +8.4 uC=+8.4 *10⁻⁶C

q₂= +5.6 uC= +5.6 *10⁻⁶C

Calculation of the distance (d) separating the charges

We replace data in the equation (1):

`0.66=(8.99*10^(9)*8.4*10^(-6) *5.6*10^(-6)  )/(d^(2) )`

`d^(2) =(422.89*10^(-3) )/(0.66)`

d²=640.74*10⁻³

`d=\sqrt{640.74*10^(-3) }`

d=0.8 m