Physics Midterm: Calculate Electric Field at Origin

[Nimmy]

Homework Statement

A two point charges (Qa and Qb) are placed along the y and x axis, as shown below:

The Qa charge is placed 3 meters up from the y axis, which contains a -5.00 microcolumb charge. The Qb charge is placed 4 meters away from the x axis, which contains a 8.00 microcolumb charge.

a. Calculate the net electric Field at the origin.

Homework Equations

E-Field on Qa = kq/r^2
E-Field on Qb = kq/r^2[c]3. Attempt of a solution[/b]E-Field on Qa = kq/r^2 = (9.0 E9 N m^2/C^2)(-5.00 E-6 C)/(3m)^2 = -5000 N/m
E-Field on Qb = kq/r^2 = (9.0 E9 N m^2C^2)(8.00 E-6 C)/(4m)^2 = 4500 N/m

Where do I go from here? Do I add the E-Fields or Find the Pythagorean Theorem to find the magnitude of the E-Field?

Thanks.

BTW...Dont worry I am not cheating...IM just asking a question that was asked on the midterm...that I didnt get the answer from.

 

[LowlyPion]

The E-field is a vector field.

So accounting for the contribution from the 2 charges means adding the vector components.

Looks like Pythagoras would be the way to go to calculate |E|.

 

[nlightner]

To find the net electric field at the origin, you will need to calculate the vector sum of the individual electric fields from Qa and Qb. This can be done using the Pythagorean theorem, as you mentioned.

First, we need to find the direction of the electric fields. The electric field from Qa will point towards the negative y-axis, since it is a negative charge. The electric field from Qb will point towards the positive x-axis, since it is a positive charge.

Next, we can calculate the magnitude of the electric fields using the equations you provided.

E-Field on Qa = 5000 N/m (since it is a negative charge, the magnitude will be positive but the direction will be negative)
E-Field on Qb = 4500 N/m (since it is a positive charge, the magnitude and direction will both be positive)

Now, we can use the Pythagorean theorem to find the magnitude of the net electric field at the origin:

E-net = √(E-Qa^2 + E-Qb^2) = √(5000^2 + 4500^2) = 6775 N/m

To find the direction of the net electric field, we can use trigonometry. The angle between the x-axis and the net electric field can be found using the inverse tangent function:

θ = tan^-1(4500/5000) = 39.8 degrees

Therefore, the net electric field at the origin is 6775 N/m, pointing at an angle of 39.8 degrees above the positive x-axis.

I hope this helps! Good luck with your midterm!