Net Electric Field at 4 cm from a 40 nC Charged Particle A - Homework Solution

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Homework Help Overview

The discussion revolves around determining the charge of particle B given that the net electric field at a point 4 cm from charged particle A (40 nC) is zero. The participants explore the implications of this condition in the context of electric fields and charge interactions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants consider the possibility of multiple charges for B that could result in a net electric field of zero at the specified point. They discuss the locations where the electric fields could cancel and question the necessary charge magnitude for B.

Discussion Status

The conversation includes various interpretations of the problem, with participants examining different scenarios for charge B. Some guidance is provided regarding the calculations needed to find the charge of B, but there is no explicit consensus on the final values or whether other solutions exist.

Contextual Notes

Participants note the importance of charge signs and distances in their calculations, as well as the implications of vector directions in electric field cancellation.

-EquinoX-
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Homework Statement


if I have two charged particles A and B and are distanced 8 cm apart. At a point 4 cm from
A, the net electric field is 0. The charge of A is 40 nC. What can we conclude about the charge of B?


Homework Equations





The Attempt at a Solution



From my point of view B can be 40nC as well.. however I am not sure if there are some other answers as well. I am pretty sure there can be more than 1 answer
 
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There are two points that are 4cm from A where the fields can cancel. They can cancel at the midpoint of A and B or they can cancel 4cm away from A on the opposite side of A from B.
 
hmm..not clear on what you meant here.. do you mind giving me a picture? how about B's charged magnitude? how large does it have to be?
 
I'm not very good at pictures. Suppose the charges are on the x-axis. Put A at x=0cm and B at x=8cm. Then the third location could be either at x=4cm or x=(-4)cm. If x=4cm, I agree with you, the charges should be equal. How about at x=(-4)cm?
 
well if it's at -4cm then the distance between A and B is only 4 cm away right? A is at 0 and B is at -4?
 
No, A stays at 0cm and B stays at 8cm. Call the third point C=(-4)cm. C is 4cm from A and 12cm from B, right? What does the charge at B have to be to cancel the E field at C?
 
I get the picture you're trying to say now. However...how do I calculate the magnitude of B based on that info??
 
Use E=k*Q/r^2. You know the distance r for each charge and you know Q for A is 40nC. Write an equation that they sum to zero and solve for the charge at B.
 
so EA + EB = 0
k*40/4^2 + k*X/12^2 = 0

and I need to solve for X?
 
  • #10
Yeah, sure. Solve for x!
 
  • #11
okay the answer I got is 360nC, is that right? and are those the only two points from A where the fields cancels?
 
  • #12
I get -360nC, don't you? The sign is important. At any other point besides those two there is an angle between the two E fields. Can two fields cancel if they are are at different angles?
 
  • #13
yes, I missed the negative signs.. as far as I know they can't cancel if it's at a different angles.. correct?
 
  • #14
Yes. Two vectors can only cancel if they are negatives of each other. So they have to point along the same line.
 
  • #15
okay.. so the conclusion is only these two, where B is 40nC and B is -360nC
 
  • #16
Seems so to me. Ok with you?
 
  • #17
Seems fine with me, I am just worried that I missed some answers.
 

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