Electric field at a point affected by two charges

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Homework Help Overview

The discussion revolves around calculating the electric field at a specific point influenced by two charges, Q1 and Q2, with given magnitudes. Participants are tasked with determining both the strength and direction of the electric field at that point.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the electric field contributions from each charge separately and then combine them. Some participants question the accuracy of the distance calculations and the values used for constants.

Discussion Status

Participants are actively discussing the calculations and potential errors, particularly regarding the distance used for Q2 and the resulting electric field strength and direction. There is acknowledgment of possible rounding issues and significant figures, but no consensus on the correctness of the method or final answers.

Contextual Notes

There are indications of potential typos in the values used for constants, and participants are encouraged to maintain precision in intermediate calculations to avoid errors. The original poster expresses uncertainty about the correctness of their results, suggesting that the problem may be flawed.

pbj_sweg
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Homework Statement


(a) What is the strength of the electric field at the position indicated by the dot?
(b) What is the direction of the electric field at the position indicated by the dot? Specify direction as an angle measured ccw from the positive x-axis.

Enhanced_2_final.jpg


##Q_{1} = 4.0 C##
##Q_{2} = 9.0 C##

Homework Equations


$$\vec{E} = \frac{Kq}{r^2}$$
$$\text{Pythagorean's Theorem:}~a^2+b^2=c^2$$
$$K = 9\times10^{-9}$$

The Attempt at a Solution


(a) ##\vec{E_{1}}## = effect of ##Q_{1}## on point
##\vec{E_{2}}## = effect of ##Q_{2}## on point

##\vec{E_{1}} = \frac{KQ_{1}}{r^2} = \frac{K(4~C)}{r^2} = \frac{K(4~C)}{0.05^2} = 1.44\times10^{13} \frac{N}{C}##
##E_{1}\hat{i} = 1.44\times10^{13} \frac{N}{C}##
##E_{1}\hat{j} = 0 ~ \frac{N}{C}##

distance between ##Q_{2}~## and the point was found with Pythagorean's Theorem to be ~0.118 m.
##\vec{E_{2}} = \frac{KQ_{2}}{r^2} = \frac{K(9~C)}{r^2} = \frac{K(9~C)}{0.118^2} = 5.8\times10^{12} \frac{N}{C}##

Breaking ##\vec{E_{2}}## into it's components:
##E_{2}\hat{i} = E_{2}(\sin(\theta)) = 2.596\times10^{12} \frac{N}{C}##
The angle is between the long leg of the triangle and the hypotenuse and was found using tan(0.05/0.1) to get 26.57°.

##E_{2}\hat{j} = E_{2}(\cos(\theta)) = 5.192\times10^{12} \frac{N}{C}##

##E\hat{i} = E_{1}\hat{i} + E_{2}\hat{i} = 1.44\times10^{13} + 2.596\times10^{12} = 1.696\times10^{13}##
##E\hat{j} = E_{1}\hat{j} + E_{2}\hat{j} = 0 + 5.192\times10^{12} = 5.192\times10^{12}##

##| \vec E | = \sqrt{(1.696\times10^{13})^{2}+(5.192\times10^{13})^{2}} = 1.77\times10^{13}##

(b) direction of electric field is ##\tan^{-1}\left(\frac{5.192\times10^{12}}{1.696\times10^{13}}\right)## = 17.01°.

Both parts of the question are incorrect, and I'm 99% sure the second part is incorrect because the components of the electric field on the point are incorrect. Could someone please point out what I've done wrong? Thank you!
 
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pbj_sweg said:
$$K = 9\times10^{-9}$$

Is this a typo? Or did you use this value?
 
Recalculate your distance value for Q2. You should keep a few extra decimal places in intermediate values such as this distance. This is to prevent rounding/truncation errors from creeping into your significant figures. Only round at the end for your final results.
 
Student100 said:
Is this a typo? Or did you use this value?
Sorry, that's a typo. It's to the 9th power. I used the correct value.
 
Last edited:
gneill said:
Recalculate your distance value for Q2. You should keep a few extra decimal places in intermediate values such as this distance. This is to prevent rounding/truncation errors from creeping into your significant figures. Only round at the end for your final results.
Correct distance should be 0.1118 meters instead. I got my final answer to be ##1.82\times10^{13}## which is still wrong. I also didn't round until my final answer. Is there something wrong with my method instead?
 
pbj_sweg said:
Correct distance should be 0.1118 meters instead. I got my final answer to be ##1.82\times10^{13}## which is still wrong. I also didn't round until my final answer. Is there something wrong with my method instead?
Your result looks good. They may be quibbling about the significant figures though.

Also, what was your angle for the field direction?

upload_2017-2-10_22-55-37.png
 
Last edited:
gneill said:
Your result looks good. They may be quibbling about the significant figures though.

Also, what was your angle for the field direction?
I initially got an angle of 17.01° but after recalculating, using the picture you provided, I got 63.25°. Both are wrong :frown:. The program usually gives some kind of notification if the issue is with sig figs. Frankly, I think my technique is correct, so maybe MasteringPhysics just has the incorrect answer?
 
Your method for calculating the angle was okay. My figure shows the original field contributions by Q1 and Q2, not the net field. I probably should have added that (let me go back and change the figure). The angle should be a bit larger than 17°, but nowhere close to 63°.
 
gneill said:
Your method for calculating the angle was okay. My figure shows the original field contributions by Q1 and Q2, not the net field. I probably should have added that (let me go back and change the figure). The angle should be a bit larger than 17°, but nowhere close to 63°.
I see. Is there anything else wrong with my calculations? Thank you so much for your help!
 
  • #10
I don't see any other problems.
 
  • #11
gneill said:
I don't see any other problems.
Thank you, seems like the problem is just incorrect.
 

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