Find the electric field at x and y coordinates?

In summary, the electric field at the point (x1 = 50 cm, y1 = 0) due to a 60 μC point charge at the origin can be calculated using Coulomb's Law and taking the magnitude of the electric field equation. After converting the charge to full Coulombs and using the distance x as the total distance from the charge, the x-component can be calculated as 2.1567*106 N/C in the positive x-direction. Since the electric field vector points in the positive x-direction, the y-component is equal to 0. A sketch can help visualize the direction and magnitude of the vector components.
  • #1
jlmccart03
175
9

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 Ex.

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!
 
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  • #2
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.
 
  • #3
gneill said:
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.
 
  • #4
jlmccart03 said:
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##.
 
  • #5
gneill said:
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?
 
  • #6
jlmccart03 said:
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.
 
  • #7
gneill said:
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.
 
  • #8
jlmccart03 said:
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 :smile: 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?
 
  • #9
gneill said:
You are definitely confusing yourself :smile: 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.
 
  • #10
jlmccart03 said:
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?
 
  • #11
gneill said:
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.
 
  • #12
jlmccart03 said:
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.
 
  • #13
gneill said:
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!
 
  • #14
Cheers! :smile:
 

1. What is the formula for calculating electric field at a given point?

The formula for calculating electric field at a given point is E = kQ/r^2, where E is the electric field strength, k is the Coulomb's constant, Q is the magnitude of the charge, and r is the distance between the charge and the point.

2. How do I determine the direction of the electric field at a given point?

The direction of the electric field at a given point is in the direction of the force that a positive test charge would experience if placed at that point. If the charge at the point is positive, the electric field points away from it, and if the charge is negative, the electric field points towards it.

3. Can the electric field at a given point be negative?

Yes, the electric field at a given point can be negative. This means that the direction of the electric field is opposite to the direction that a positive test charge would experience. It is important to note that the electric field itself is a vector quantity, and its magnitude and direction can vary at different points.

4. How does the distance affect the magnitude of the electric field at a given point?

The magnitude of the electric field at a given point is inversely proportional to the square of the distance between the point and the charge. This means that as the distance increases, the electric field decreases. This relationship is described by the inverse square law.

5. Is the electric field at a given point affected by the presence of other charges?

Yes, the electric field at a given point can be affected by the presence of other charges. This is because electric fields can interact with each other, and the resulting field at a given point can be a combination of the individual fields from each charge. This is known as the principle of superposition.

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