# Electric fields

• Mar 4th 2007, 10:04 AM
totalnewbie
Electric fields
Charges in the corner are 10*10^(-9) C.
What is strenght of electric field at red point.
• Mar 4th 2007, 01:58 PM
topsquark
Quote:

Originally Posted by totalnewbie
Charges in the corner are 10*10^(-9) C.
What is strenght of electric field at red point.

You need to find the electric field at that point due to each charge then add them vectorally.

Call q = 10 nanoCoulomb and r = 0.05 m
I am setting up a cooredinate system with +x to the right and +y upward.

So:
E(A) = k*q(A)/r^2(A) = k*q/(r/2)^2 = 4kq/r^2
(If you aren't using "k" just substitute k = 1/(4*pi*epsilon0) )

This field will be pointing away from A, so this E component is pointing in the -y direction.

E(D) = k*q(D)/r^2(D) = k*q/(r/2)^2 = 4kq/r^2

This field will be pointing toward D, so this E component is pointing in the -y direction.

E(B) = k*q(B)/r^2(B) = k*q/(r^2 + (r/2)^2) <-- By the Pythagorean theorem

E(B) = 4kq/5r^2

This field will be pointing toward B, so
E(B)x = E(B)*cos(t)
E(B)y = E(B)*sin(t)
where
sin(t) = (r/2)/sqrt{5r^2/4} = 1/sqrt{5}
cos(t) = r/sqrt{5r^2/4} = 2/sqrt{5}

So
E(B)x = 2E(B)/sqrt{5} = 8kq/(5sqrt{5}*r^2)
E(B)y = E(B)*sin(t) = 4kq/(5sqrt{5}*r^2)

Similarly:
E(C)x = -8kq/(5sqrt{5}*r^2)
E(C)y = 4kq/(5sqrt{5}*r^2)

So the overall electric field E has components:
Ex = 0 + 8kq/(5sqrt{5}*r^2) + -8kq/(5sqrt{5}*r^2) + 0 = 0

Ey = -4kq/r^2 + 4kq/(5sqrt{5}*r^2) + 4kq/(5sqrt{5}*r^2) + -4kq/r^2 = - 8kq/r^2 + 8kq/(5sqrt{5}*r^2)

The magnitude of E at this point is then:
E = sqrt{Ex^2 + Ey^2} = sqrt{[0]^2 + [- 8kq/r^2 + 8kq/(5sqrt{5}*r^2)]^2} = - 8kq/r^2 + 8kq/(5sqrt{5}*r^2)

-Dan