Electric Field Video Lessons

Concept

# Problem: A point charge q1 = -4.00 nC is at the point x = 0.600 meters, y = 0.800 meters, and a second point charge q2 = +6.00 nC is at the point x = 0.600 meters, y = 0. A) Calculate the magnitude E of the net electric field at the origin due to these two point charges.B) What is the direction, relative to the negative x-axis, of the net electric field at the origin due to these two point charges?

###### FREE Expert Solution

Electric field due to point charge:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{k}\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}}$

Vector magnitude and direction:

$\overline{)\mathbf{|}\stackrel{\mathbf{⇀}}{\mathbf{r}}\mathbf{|}{\mathbf{=}}\sqrt{{{\mathbf{r}}_{\mathbf{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{r}}_{\mathbf{y}}}^{\mathbf{2}}}}$

$\overline{){\mathbf{tan}}{\mathbf{\theta }}{\mathbf{=}}\frac{{\mathbit{r}}_{\mathbf{y}}}{{\mathbit{r}}_{\mathbf{x}}}}$

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###### Problem Details

A point charge q1 = -4.00 nC is at the point x = 0.600 meters, y = 0.800 meters, and a second point charge q2 = +6.00 nC is at the point x = 0.600 meters, y = 0.

A) Calculate the magnitude E of the net electric field at the origin due to these two point charges.

B) What is the direction, relative to the negative x-axis, of the net electric field at the origin due to these two point charges?