Find the electric field at x and y coordinates?

[jlmccart03]

Homework Statement

A 60 μC point charge is at the origin.
Find the electric field at the point x1 = 50 cm , y1 = 0.

Homework Equations

Coulombs Law: F = (kQ₁Q₂)/r²
Electric Field: E = kQ/r²
k = 8.99*10⁹ C

The Attempt at a Solution

So taking Coulombs Law and deriving the Electric Field equation I then took my charge particle and set that equal to Q (First converting it from micro to full Coulombs), then taking the distance x and setting that equal to the r² I put that in the denominator to finally solve for the E_x.

Note that k is 8.99*10⁹

Finally I do the same thing with the y-component, but this is my issue. When I take the numbers and divide by 0 I get the infamous infinity.

So What Am I doing wrong here?

Thanks in advance!

 

[gneill]

The r in the denominator is the total distance from the charge, and includes the contributions of both x and y offsets at the same time. This version of the Coulomb's law equation is a scalar version, yielding the magnitude of the electric field. It's up to you to assign the direction (i.e. break it into x and y components) as required.

 

[jlmccart03]

The r in the denominator is the total distance from the charge, and includes the contributions of both x and y offsets at the same time. This version of the Coulomb's law equation is a scalar version, yielding the magnitude of the electric field. It's up to you to assign the direction (i.e. break it into x and y components) as required.

Wait so the answer I get when taking .5+.0 and squaring it is the same answer for both the x and y components? This is where I am starting to get confused.

 

[gneill]

Wait so the answer I get when taking .5+.0 and squaring it is the same answer for both the x and y components? This is where I am starting to get confused.

Yes. But think Pythagoras: $r^2 = x^2 + y^2$.

 

[jlmccart03]

Yes. But think Pythagoras: $r^2 = x^2 + y^2$.

Ok so r² = (0.5)²+0² = 0.25
Then taking this I simply get the equation (8.99*10⁹*6*10^-5)/0.25 = 2.1576*10⁶.
So I guess I am confused again on how I differentiate between x and y components. Is it the same answer for both?

 

[gneill]

So I guess I am confused again on how I differentiate between x and y components. Is it the same answer for both?

What you have calculated is the magnitude of the electric field (be sure to include the units!). You need to break that into x and y components depending upon the direction of the vector it's associated with. This case is pretty simple given the direction of the electric field vector at the location you were given.

 

[jlmccart03]

What you have calculated is the magnitude of the electric field (be sure to include the units!). You need to break that into x and y components depending upon the direction of the vector it's associated with. This case is pretty simple given the direction of the electric field vector at the location you were given.

Ok, so I get 2.1567*10⁶ N/C and since initially x = 50 cm or 0.5 m and y = 0 cm or 0 m then my x-component is 2.1567*10⁶ N/C in the positive x-direction, and 2.1567*10⁶ N/C in the positive y-direction? I might be confusing myself.

 

[gneill]

Ok, so I get 2.1567*10⁶ N/C and since initially x = 50 cm or 0.5 m and y = 0 cm or 0 m then my x-component is 2.1567*10⁶ N/m in the positive x-direction, and 2.1567*10⁶ in the positive y-direction? I might be confusing myself.

You are definitely confusing yourself If you were to sketch the E vector on a diagram that shows the charge and the location in question on XY axes, what direction would it point?

 

[jlmccart03]

You are definitely confusing yourself If you were to sketch the E vector on a diagram that shows the charge and the location in question on XY axes, what direction would it point?

It would point to the right. Correct? The charge is positive and the E vector is positive so they would be repulsive.

 

[gneill]

It would point to the right correct since the charge is positive and the E vector is positive they would be repulsive.

Well, the E vector doesn't have a sign per se, but yes, it would point to the right in the +x direction. What does that tell you about the magnitude of the y-component for such a vector?

 

[jlmccart03]

Well, the E vector doesn't have a sign per se, but yes, it would point to the right in the +x direction. What does that tell you about the magnitude of the y-component for such a vector?

The y vector magnitude is 0 since it does not point in either direction on the y-axis.

 

[gneill]

The y vector magnitude is 0 since it does not point in either direction on the y-axis.

Correct. So the x-component must account for the entire magnitude of the vector.

 

[jlmccart03]

Correct. So the x-component must account for the entire magnitude of the vector.

Thank you so much for sticking with me! That was my entire issue was the y-component and once I drew a sketch then it just pieced together. Thanks again!

 

[gneill]

Cheers!