Electric Field and Force from four particles

Four charged particles are arranged in a square as shown below, with A=3, B=4, and C=5

(a) Determine the electric field at the location of charge q. (Use the following as necessary: $q$, $a$, and $k_e$.) Both the magnitude and the direction, with direction given in degrees counterclockwise from the x-axis.

(b) Determine the total electric force exerted on q. (Use the following as necessary: $q$, $a$, and $k_e$.) Again, direction and magnitude.

The work I have so far:

I know that $E=\frac{k_eq}{r^2}$, so I figured that $E_A=\frac{3qk_e}{a^2}$ and $E_C=\frac{5qk_e}{a^2}$ and $E_B=\frac{4qk_e}{2a^2}$ since the distance between B and the upper right corner is $\sqrt{2a^2}$, and I think I have to split $E_B$ into the x and y components to add them, but how does that give me a single magnitude and not a combination of x and y parts? I honestly have no idea how to find the direction of the field.

part b should be similar, except with $F_e=\frac{k_eq_1a_2}{r^2}$ for each one, which I have worked out, and then added similarly with B split into x and y components, right? Again, not sure the formula or trick to find the direction of the force.

• This might be the confusing part: how do you get the x and y components of $E_B$? Commented Jun 8, 2016 at 22:07
• @philip_0008, wouldn't the x be $E_Bcos(45)$ and y be $E_Bsin(45)$? Commented Jun 8, 2016 at 22:10
• Ah you're right. Even that confuses me. I've just deleted an answer saying use $a$ for the distance of x-component:) Commented Jun 8, 2016 at 22:12

If you found the total x and y component of the total field just use:

Magnitude E_{tot}:

$$E_{tot}=sqrt(E^2_{tot-x}+E^2_{tot-y})$$

The direction in angle:

$$\theta = arctan(\frac{E_{tot-y}}{E_{tot-x}})$$

The magnitude of force would just be $F=qE$

Note: you should be careful in angles as it may sometimes not the angle measured from +x as the norm, for you to detmine it, check at which quadrant your resultant vector lie, Its for you to figure out (Trigo basics).

a) Add the x and y components separately, then calculate the magnitude E (the resultant field) and the angle it makes with the x axis.

b) The force is qE, in the same direction as E.