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# Calculate the Magnitude and Direction of the NET Electric Field?

I've included a little sketch just to help visualize the orientation of electric fields and distance.

The length of each grid is represented by 0.1m. q1 (blue dot) = - 4.0 x 10^-6 C and q2 (red dot) = + 2.0 x 10^-6 C.

Can I have help calculating the magnitude of the net electric field on the black dot?

I'm sort of struggling with my trigonometry and vectors. I know that the angles are 45 degrees based on the triangles within the figure and the distance from each charge to C, but am struggling on where to go from here.

### 1 Answer

- 11 months agoFavorite Answer
The electric field (E) due to a point charge (Q) at a point at a distance (r) is given by:

E = kQ / r²

k is the coulomb constant = 8.99×10⁹ N·m²·C⁻²

The field is away from a positive point charge and towards a negative point charge.

{Remember the way to test a field is to put a unit positive test charge at the point in question and see which way the resulting force acts on it.}

So we need the diagonal distances (r) from the red to black and blue to black.

Uncle Pythagoras:

r² = 0.2² + 0.2²

r² = 0.08

E due to red dot = (8.99×10^9 * 2.0×10^-6 / 0.08) = 2.25 ×10^5 V/m { to the north west :}

Similarly

E due to blue dot = (8.99×10^9 * 4.0×10^-6 / 0.08) = 4.50 ×10^5 V/m { to the south west :}

The resultant field at the black dot is just the vector sum of the two field components {purple}.

magnitude from Pythagoras

R = 5.03 ×10^5 V/m

direction = arctan (2.25/4.5) = 26.6 degrees north of south west = 18.4 degrees south of west

{So invent your own coordinate system then! :-) }