PLEASE help! PHYSICS homework question!?

Question

Charges of 4x10^-6 and -6x10^-6 are placed at two corners of an equilateral triangle with sides of 0.10m. At the third corner, what is the electric field magnitude created by these two charges? The answer is 4.8x10^6 N/C but I want to know how! Thanks!

Answer 1

Electric field due to a charge = 9*10^9(Q/r).

The magnitudes of the fields due to these charges are:

9*10^9(4x10^-6/0.1) = 36*10^4 = 3.6*10^5

9*10^9(-6x10^-6/0.1) = 54*10^4 = 5.4*10^5.

Add these two fields vectorially. They are at a 60 degree angle from each other. One could use the law of cosines.

a^2 = b^2 + c^2 - 2bc cosA

a^2 = (3.6^2 + 5.4^2 -2*3.6*5.4*0.5)*10^10

a^2 = 22.68*10^10

a = 4.76*10^5

The given answer is off by a factor of 10.

One could also resolve one of the vectors into components parallel and perpendicular to the other. For instance, resolve the field due to the 4x10^-6 charge into components parallel and perpendicular to the -6x10^-6 charge.

Take the direction of the field from the -6x10^-6 charge to be the negative x direction. The field due to the 4x10^-6 charge in the x direction is 9*10^9(4x10^-6/0.1)cos60.

The field due to the 4x10^-6 charge in the y direction is 9*10^9(4x10^-6/0.1)sin60 = 3.12*10^5.

The total field in the x direction is -5.4*10^5+1.8*10^5= - 3.6*10^5.

Use Pythagoras to add the two components and get the same answer.

EDIT:

Helmut is right. You have to square r. That accounts for my answer being off by a factor of ten.

Answer 2

F = (8.987551787e+9 Nm^2/C^2)q1q2/r^2

E = (8.987551787e+9 Nm^2/C^2)Q/r^2

E1 = (8.987551787e+9 N/C)(4e-6) / (0.01 m^2) @ 0°

E2 = (8.987551787e+9 N/C)(- 6e-6) / (0.01 m^2) @ 60°

E1 ≈ 3.595021e+6 N/C @ 0°

E2 ≈ - 5.392531e+6 N/C @ 60°

E2 ≈ 5.392531e+6 N/C @ - 60°

Using the Law of Cosines for the vector sum,

|E| ≈ √((3.595020e+6)^2 + (5.392531e+6)^2 - 2(3.595021e+6)(5.392531e+6)cos(- 60°))

|E| ≈ (1e+6)√(12.924173939841 + 29.079391364642 - 19.386260909762)

|E| ≈ 4.755765e+6 N/C ≈ 4.8e+6 N/C