Find the magnitude of the electric field at the orgin due to these three charges. In units of N/C?

Question

Three equal charges 4.3 uC are located in the xy-plane, one at (0 m, 31 m), another at (69 m, 0 m), and the third at (60 m, -62 m).

i have the equation but I dont know the x and y components.

Answer 1

We generally express electric field in terms of Newtons/Coulomb, so the real problem here is determining the electric field.

Remember: E = k*q/r^2 where k = Coulomb's constant = 8.988e9, q = source charge, and r = distance between source charge and point of reference.

Now, this gives you the magnitude of the electric field due to a single point charge. If you multiply this magnitude by the unit vector pointing from your point of reference (in this case the origin) to the point charge, you will get the electric field as a vector. Then, all you'll have to do is add up the x and y components due to each of the three charges to obtain the net electric field.

In case you've forgotten, the unit vector is just the vector divided by the magnitude of the vector. For your first and second charges, the unit vector is <0,1> and <1,0>, respectively. The sum of the squares of a unit vector should always equal 1.

To solve for the third unit vector:

The vector (not normalized, therefore not a unit vector) pointing towards the third charge is <60, -62>. However, considering the aforementioned stipulations of a unit vector, we see that we must divide this by the magnitude of the vector. We find the magnitude using the Pythagorean Theorem, where the magnitude = sqrt(60^2 + 62^2) = 86.279. Thus, our unit vector is <60/86.279 , -62/86.279> = <.695, -.719>. Notice that .695^2 + .719^2 = 1, therefore this is a unit vector! Of course, these numbers I'm using are approximations and your teacher may want you to be a little more precise.

Now, all you have to do is multiply the magnitude by the x and y components of your vectors, then sum up the x and y components of your electric fields to get the net electric field in terms of x ( î ) and y ( ĵ ).

And remember, electric fields always "point toward" negative charges. Because all of these charges are positive, the electric field should "point" away from them.

Answer 2

How would i'm getting ? if the two actual one among my expenses are 4 micro C placed at x= 1m and x =-1m and that i'd opt to calc the fee E of the internet electric powered field at y= 0.6m? idk what to anticipate x and y to plug into arctan?, the only way i think of like i will do it finally finally end up being 0 via utilising certainty it may nicely be arctan(0/0.6)