What is the Electric Field at Two Points?

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Homework Help Overview

The discussion revolves around calculating the electric field generated by a point charge located at the origin, specifically at the coordinates x3 = -35 cm and y3 = 80 cm. Participants are exploring the components of the electric field and the appropriate use of trigonometric functions in their calculations.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the electric field's components using trigonometric functions based on the geometry of the problem. There is confusion regarding the correct application of sine and cosine, particularly in relation to the distances involved and the angle to the x-axis. Questions arise about the assumptions made in the calculations and the interpretation of the signs of the components.

Discussion Status

Several participants have provided feedback on the original poster's calculations, prompting a reevaluation of the trigonometric ratios used. There is ongoing clarification regarding the correct values for the components and the overall approach to finding the electric field. Some participants suggest rechecking calculations for accuracy, while others emphasize the importance of understanding the geometric relationships in the problem.

Contextual Notes

The original poster is working under constraints of limited attempts to submit their answers, which adds pressure to ensure accuracy in their calculations. There is also a focus on the correct signs of the components based on the charge's position and the coordinate system used.

jlmccart03
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Homework Statement


A 90 μC point charge is at the origin.
Find the electric field at the point x3 = -35 cm , y3 = 80 cm .
Express your answers using two significant figures. Enter the x and y components of the electric field separated by a comma.

Homework Equations


E = kQ/r^2
magnitude * cos/sin of theta

The Attempt at a Solution


So I found the Etot to be 1.06*106 N/C and from there I drew a picture and found each component by using sin and cos of theta so I get Ex = 1.06*106*cos(.35/.7625) and 1.06*106*cos(.8/.7625) for the Ey. This is wrong however. So what steps am I missing here?
 
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How did you arrive at the cos and sin arguments? I don't see where you've accounted for the sign of the x3 value in your component calculations; do you expect the field vector to have two positive components? What does a look at your sketch of the scenario suggest? (you did make a sketch first, right?)
 
gneill said:
How did you arrive at the cos and sin arguments? I don't see where you've accounted for the sign of the x3 value in your component calculations; do you expect the field vector to have two positive components? What does a look at your sketch of the scenario suggest? (you did make a sketch first, right?)
This is all my work. I found the total E field magnitude, but I need it in components. So originally I just did it in components with each x and y value. This I realized is wrong. Both x and y contribute to the total r in the E field equation for a point charge. I then decided ok, I have magnitude, but need components not net. I then thought that I could simply take the magnitude E and multiply by the corresponding angle using sin or cos. Is sin or cos not even used?
 

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jlmccart03 said:
This is all my work. I found the total E field magnitude, but I need it in components. So originally I just did it in components with each x and y value. This I realized is wrong. Both x and y contribute to the total r in the E field equation for a point charge. I then decided ok, I have magnitude, but need components not net. I then thought that I could simply take the magnitude E and multiply by the corresponding angle using sin or cos. Is sin or cos not even used?
You seem to have a basic misunderstanding of trigonometry.
The sine and cosine are functions of the angle, but the angle is not the ratio between the two distances.
In fact, you do not need to find the angle, since you have the distances and the sine and cosine values come directly from their ratios.
For the x component, the cosine of the angle of r to the x-axis is just the x distance divided by r.
(Where did you get 0.7625 from? r must be more than 0.8.)
 
haruspex said:
You seem to have a basic misunderstanding of trigonometry.
The sine and cosine are functions of the angle, but the angle is not the ratio between the two distances.
In fact, you do not need to find the angle, since you have the distances and the sine and cosine values come directly from their ratios.
For the x component, the cosine of the angle of r to the x-axis is just the x distance divided by r.
(Where did you get 0.7625 from? r must be more than 0.8.)
0.7625 is r. I just did (0.352+0.82) and then took the square root of that which gave 0.7625. So its cos(.35/r) for x?
 
jlmccart03 said:
then took the square root
No, it looks like you forgot to do that. Look at the triangle. How could r be less than y?
jlmccart03 said:
So its cos(.35/r) for x
No. If θ is the angle r makes to the x-axis then cos(θ)=.35/r.
 
haruspex said:
No, it looks like you forgot to do that. Look at the triangle. How could r be less than y?

No. If θ is the angle r makes to the x-axis then cos(θ)=.35/r.
I see so r is 0.8732 and cos(θ)=.35/.8732 = .400819. Then multiply by E to get x-component?
 
jlmccart03 said:
I see so r is 0.8732 and cos(θ)=.35/.8732 = .400819. Then multiply by E to get x-component?
Yes.
 
haruspex said:
Yes.
Ok so I get for the x-component 4.3*105 N/C and for the y-component I get 9.7*105 N?C. I am asking since I only have one attempt left. I took the 1.06*106 * (0.35/0.8732) for the x-component and 1.06*106 * (0.80/0.8732) to get the two above numbers.
 
  • #10
jlmccart03 said:
Ok so I get for the x-component 4.3*105 N/C and for the y-component I get 9.7*105 N?C. I am asking since I only have one attempt left. I took the 1.06*106 * (0.35/0.8732) for the x-component and 1.06*106 * (0.80/0.8732) to get the two above numbers.
That's essentially correct, but I make the x component just under 4.25, so closer to 4.2 than 4.3. Not sure how accurate you need to be. I recommend you run the calculations again keeping one more digit of accuracy until the end. I can believe your 4.3 is right and I am wrong.
 
  • #11
jlmccart03 said:
Ok so I get for the x-component 4.3*105 N/C and for the y-component I get 9.7*105 N?C. I am asking since I only have one attempt left. I took the 1.06*106 * (0.35/0.8732) for the x-component and 1.06*106 * (0.80/0.8732) to get the two above numbers.
What is the correct sign of Ex?
 
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  • #12
ehild said:
What is the correct sign of Ex?
It is negative. Thanks for the help everyone!
 

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