Finding the magnitude of an electric field with 5 charges.

  • Thread starter Ganar
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1. Five charges of 1.00 μC magnitude are placed in a row, each charge being 1.00 m distant from its neighboring charges. If all chargers are positive, what is the magnitude in N/C, of the electric field at the position of charge 4?
A) 0
B)2.13x10^-2
C)3.25x10^-3
D)3.25x10^3
E)2.13x10^4


2. Homework Equations :
kq/r^2



3. I used this equation yet it doesn't yield any of the solutions so I'm wondering if there is an extra step and if the fact that they are all positively charged effects the process of solving this type of problem. Help being pointed in the right direction would be helpful and equations used would be greatly appreciated.
 

Answers and Replies

  • #2
collinsmark
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Hello Ganar,

Welcome to Physics Forums!
1. Five charges of 1.00 μC magnitude are placed in a row, each charge being 1.00 m distant from its neighboring charges. If all chargers are positive, what is the magnitude in N/C, of the electric field at the position of charge 4?
A) 0
B)2.13x10^-2
C)3.25x10^-3
D)3.25x10^3
E)2.13x10^4

2. Homework Equations :
kq/r^2

3. I used this equation yet it doesn't yield any of the solutions so I'm wondering if there is an extra step and if the fact that they are all positively charged effects the process of solving this type of problem. Help being pointed in the right direction would be helpful and equations used would be greatly appreciated.

Don't forget that the electric field is a vector. It has both magnitude and direction.

When you add together multiple electric fields using the superposition principle, you must add them together as vectors. :wink:
 
  • #3
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So that means I would get a set of equations like so:
F14= k(q1q4)/(3-x)^2
F24=k(q2q4)/(2-x)^2
F34=k(q3q4)/(x)^2
F54=k(q5q4)/(x)^2
And I set these all equal to each other and will end up with a quadratic expression then use the quadratic formula to solve the problem?
 
  • #4
collinsmark
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So that means I would get a set of equations like so:
F14= k(q1q4)/(3-x)^2
F24=k(q2q4)/(2-x)^2
F34=k(q3q4)/(x)^2
F54=k(q5q4)/(x)^2
That's pretty much the general form of each constituent force. But you're forgetting about the direction. Force, like electric field, is a vector. For example, is q2 pushing q4 to the right or to the left? Which direction does q5 push q4?

And since you're tasked with finding the electric field at the point (not the force), so you'll have to remove one of the variables from the above equations*.

*(Personally, I think the problem would have made more sense asking you for the force rather than the electric field. It arguably makes things less clear asking for the electric field at the point where q4 is, when q4 is still there. So treat this problem as q4 being removed, and as if asking for the electric field at the point where q4 would have been if it were still there.)
And I set these all equal to each other and will end up with a quadratic expression then use the quadratic formula to solve the problem?
No, you don't set them all equal to each other.

The superposition principle states the the net electric field is the vector sum of the individual constituent electric fields. You need to sum them. But you must sum them as vectors, taking their direction into account.
 
  • #5
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Alright thank you I'll keep working at it if I have anymore questions I'll go ahead and ask them. Thanks again for the help!!
 

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