Calculating the electric field, multiple point charges

Etotal should be (-96250, 92272) for EtotalIn summary, the electric field at the position indicated by the dot is (-96250, 92272) N/C, with a strength of 133534 N/C. The horizontal component is negative and the vertical component is positive, indicating that the field is pulling the dot towards the negative charge located at the position (-0.02m, 0.04m). The equation used to calculate this was [1/(4πε0)]*(|q|/r2), where q is the charge and r is the distance between the charges. The calculations involved converting units and taking into account the positive
  • #1
aliaze1
174
1

Homework Statement



What are the strength and direction of the electric field at the position indicated by the dot?

knight_Figure_26_32.jpg


Part A:
Give your answer in terms of the horizontal and vertical components, separated by commas. Take the positive directions to be up and to the right.

Part B:
Specify the strength of the electric field.

Homework Equations



[1/(4πε0)]*(|q|/r2)

The Attempt at a Solution



[1/(4πε0)]=9*109

q1=-5*10-9
q2=10*10-9
q3=10*10-9

r1=0.02m
r2=0.04m
r3=√(0.002)=(0.002)0.5=~0.0447m

E1={[9*109]*[5*10-9]}/(0.02m)2
E2={[9*109]*[10*10-9]}/(0.04m)2
E3={[9*109]*[10*10-9]}/(0.002m)

E1=112500 j
E2=56250 i
E3=45000

E3_i=45000*(0.04/0.0447)=40249 i
E3_j=45000*(0.02/0.0447)=20134 j

Right is positive, therefore:
E2= -56250 i
E3_i= -40249 i

ETotal_i= -56250 + -40249 = -96499 i
ETotal_j= 20134 + 112500 = 133534 j

This is coming out as 'incorrect', and therefore I cannot proceed to part 2

Where am I going wrong?

Thanks!
 
Last edited:
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  • #2
You need to redo E3.
 
  • #3
neutrino said:
You need to redo E3.

typo, sorry, E3 is actually 45000, unless you mean that that is also incorrect
 
Last edited:
  • #4
Oh, okay. 45000 is correct.

I just calculated the x-comp. of E3 and it comes out to 40,268.45638, while your answer is off by almost 20 N/C. I think it's a calculation error somewhere.
 
  • #5
neutrino said:
Oh, okay. 45000 is correct.

I just calculated the x-comp. of E3 and it comes out to 40,268.45638, while your answer is off by almost 20 N/C. I think it's a calculation error somewhere.

yea idk for some reason my calculator is giving me the 40249.22359..., anyways, is my y-component correct? i have very few attempts left

thanks
 
  • #8
...still isn't working :(
 
  • #9
aliaze1 said:

Homework Statement



What are the strength and direction of the electric field at the position indicated by the dot?

knight_Figure_26_32.jpg


Part A:
Give your answer in terms of the horizontal and vertical components, separated by commas. Take the positive directions to be up and to the right.

Part B:
Specify the strength of the electric field.

Homework Equations



[1/(4πε0)]*(|q|/r2)

The Attempt at a Solution



[1/(4πε0)]=9*109

q1=-5*10-9
q2=10*10-9
q3=10*10-9

r1=0.02m
r2=0.04m
r3=√(0.002)=(0.002)0.5=~0.0447m // r3 = 0.4472

E1={[9*109]*[5*10-9]}/(0.02m)2
E2={[9*109]*[10*10-9]}/(0.04m)2
E3={[9*109]*[10*10-9]}/(0.002m)

E1=112500 j
E2=56250 i
E3=45000

E3_i=45000*(0.04/0.0447)=40249 i
E3_j=45000*(0.02/0.0447)=20134 j

Right is positive, therefore:
E2= -56250 i
E3_i= -40249 i

ETotal_i= -56250 + -40249 = -96499 i
ETotal_j= 20134 + 112500 = 133534 j

This is coming out as 'incorrect', and therefore I cannot proceed to part 2

Where am I going wrong?

Thanks!

The answer is E(0,0.2) = (-964.15,-922.9) where E(0,0.2) = 1334.7
 
Last edited:
  • #10
please correct the value of r3, then everything should fall into place.
 
  • #11
I got (-96392,-92272)

your J value is incorrect. your e1 is incorrect. using the correct sign convention, since e1 is a negative charge, it pulls the dot towards it. not away, therefore e1 should be negative since up and right is postive
 

1. How can I calculate the electric field of multiple point charges?

To calculate the electric field of multiple point charges, you can use the principle of superposition. This means that you calculate the electric field created by each individual charge and then add them together to get the total electric field at a certain point. The formula for this calculation is E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the point charge, and r is the distance between the point charge and the point where you want to calculate the electric field.

2. Can the electric field of two point charges cancel each other out?

Yes, it is possible for the electric field of two point charges to cancel each other out at a certain point. This is known as a neutral point. The electric field at this point would be zero because the forces from the two point charges would be equal in magnitude and opposite in direction, resulting in a net force of zero.

3. How does the distance between point charges affect the electric field?

The electric field is inversely proportional to the square of the distance between the point charges. This means that as the distance increases, the electric field decreases. This relationship is described by the inverse square law, which states that the intensity of a physical quantity is inversely proportional to the square of the distance from the source of that quantity.

4. Can the direction of the electric field change for multiple point charges?

Yes, the direction of the electric field can change when dealing with multiple point charges. This is because the electric fields from each individual point charge can have different magnitudes and directions. When adding them together, the resulting electric field could have a different direction than each individual field.

5. How can I determine the net electric field at a certain point due to multiple point charges?

To determine the net electric field at a certain point due to multiple point charges, you can use vector addition. This means that you must take into account the magnitude and direction of each individual electric field and use the rules of vector addition to find the resulting net electric field. This calculation can also be done using the principle of superposition, as mentioned in the first question.

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