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?

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caledon | High School Teacher | (Level 3) Senior Educator

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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

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