Question

A point charge of `3.00×10^-6 C` is `12.0cm` distant from a second point charge of `−1.50×10^-6 C`. Calculate the magnitude of the force on each charge?

Answer 1

`2.813" N"`

Explanation:

Write out all the givens for this problem.

`"Given"`
`q_(1) = 3.00xx10^(−6)C`
`q_(2) = -1.50xx10^(-6)C`
`r = 12.0" cm"or(0.12" m")`

To calculate the Coulombic Force between 2 point charges, you use the following:

`color(white)(aaaaaaaaaaaaaaaa)color(magenta)(F_(c) = (k |q_(1)|q_(2)|)/r^2`

`color(white)(aaaaaaaaaaaaa)` `"Since we are only looking for the"`
`color(white)(aaaaaaaaaaaaa)` `"magnitude of the force, we use"`
`color(white)(aaaaaaaaaaaaa)` `"the absolute values of each point"`
`color(white)(aaaaaaaaaaaaa)` `"charge, so the signs are ignored"`

Where
`k = "Coulomb's constant" (9xx10^9" (N*m^2)/(C^2))`

`---------------------`

Plugin

`color(magenta)(F_(c) = (k |q_(1)|q_(2)|)/r^2`

`F_(c) = [(9*10^9" (N*cancelm^2)/(cancel(C^2))) * (3.00*10^(−6)cancelC)*(1.50*10^(-6)cancelC)]/(0.12cancelm)^2`

`color(blue)(|F_(c)|= 2.813" N"`

Note: If we included the signs in the formula, a (-) force would indicate attraction between the two charges and a (+) force would indicate a repulsion.

`color(white)(aaa)`

The force on `q_(1)` due to `q_(2)` is `2.813" N"` and the force on `q_(2)` due to `q_(1)` is also `2.813" N"`. Magnitude of the force is the same but the direction is opposite.

#color(white)(aaaaaaaaa)#![http://www.calctool.org/CALC/phys/electromagnetism/point_charges]()