# Electric field and charges

• jwl322

## Homework Statement

Two point charges are on the y-axis. One, with charge a of +q, is located at point a. The other, with a charge of +2q is located at point -a. Determine the magnitude and direction of the electric field at the origin. Express answers in terms of q,a, and constants.

E=kq/r^2

## The Attempt at a Solution

Do i just do k(2q)/a^2 - k(q)/a^2 and that's the answer?

K(2q)/a^2 - k(q)/a^2= kq/a^2

this in preparation of the ap exam and i learned this last semester so i just need a little refreshing

Yes, and the direction is the direction a positive test charge would go

Yup, that's how I would do it.

Since both are a distance "a" from the origin and we only care about what happens at the origin, you equations makes sense to me.

EDIT: oops, turdferguson beat me to it.

the direction is in the -y direction right?

also, if i were to find the electric potential at the origin i would use E=V/d? if i use that, I am not sure what i would use for d

thanks

Think about the direction again, all charges are positive

d is just the distance a. A good thing to remember for the AP is that fields are vectors and potentials are scalars. Voltage is also the scalar sum of each kq/r

ok so its in the positive y direction because the the electric field points outward from positive charges and since the 2q charge is greater the field at the origin will point in the positive y direction