A 75 nC charge is 8.0 m from a second charge, and the force between them is 1.0 N.  What is the magnitude of the second charge?



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Posted on (Answer #1)

Coloumb's law tells us that the electrostatic force of attraction between point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance (hence, it is also known as Coloumb's inverse square law).

The formula for force is given by:

`F = k (|q_1 q_2|)/r^2.`

k is Coloumb's Law constant (`8.99 x 10^9 (Nm^2)/(C^2)` ), q correspond to the charges (unit is C) and r the distance between the point charges (in meters).

To calculate for `q_2` :

`q_2 = (Fr^2)/(kq_1).`

F = 1.0 N

q1 = 75nC = 7.5 x 10^-8 C

r = 8.0 m


`q_2 = (1.0*8.0^2)/(8.99x10^9 * 7.5 x 10^(-8)).` 

This gives us: 0.095 Coloumbs


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