The electric field due to a point charge at a distance 6m from it is 630 N/C. What is the magnitude of the charge?
Recall that electric fields are affected by an inverse-square function; as distance from the charge increases in a linear fashion, the field diminishes exponentially. The equation to express this is:
` E = (kq)/(d^2)`
` ` where E is the strength of the field, k is Coulomb's constant (9e9), q is the charge and d-squared is the distance.
First we need to rearrange the equation. We know the following:
E = 630 N/C
k = 9.0 x 10e9
d = 6 m
So we need to solve for q. Multiply both sides by d-squared, and divide by k, to isolate q.
` (E d^2)/k = q`
Then just plug in the values we know to find the answer;
` <br data-mce-bogus="1"> `
`(22680)/(9 * 10^9)` = .00000252 C
q = 2.52 x 10e-6