Charges in the corner are 10*10^(-9) C.

What is strenght of electric field at red point.

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- March 4th 2007, 11:04 AMtotalnewbieElectric fields
Charges in the corner are 10*10^(-9) C.

What is strenght of electric field at red point. - March 4th 2007, 02:58 PMtopsquark
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