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Electric field at point?

  • Thread starter danish1991
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  • #1
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Homework Statement


Two fixed charges, A and B, are located at x axis. A is at x=0m, B is at x=4m. Qa=+4.0uc and Qb=+4.0uc. Calculate the electric field at point x=1m,2m, 3m.



Homework Equations


Qa=+4.0x10^-6 Qb=4.0x10^-6
E=F/Q-> E=(k)(q)/r



The Attempt at a Solution


Can someone help me out with this equation? I haven't been introduced to 3point distances yet. So i don't know how you would set it up, can someone walk me through this? Thanks!
 

Answers and Replies

  • #2
gneill
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It is always a good idea to draw a diagram to begin with. Pencil in the directions of the electric fields caused by each charge in the places of interest. That will help guide you when it comes time to sum up their contributions at a given point (whether to add or subtract the magnitudes).

Calculate the contributions of each charge separately, one at a time, and then sum the results (this is allowed because electric fields obey the principle of superposition, which you should have learned about as a being a property of linear systems).

Regarding your Relevant Equations, (k)(q)/r gives the electric potential (in Volts) at distance r from a charge q. You're looking for the electric field strength, which has slightly different units: Volts per meter (or equivalently, Newtons per Coulomb). Change the 'r' to 'r2' in your formula.
 
  • #3
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Okay, but what would i do with the 3 distances mentioned in the last distance, 1m,2m,3m?
 
  • #4
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It is always a good idea to draw a diagram to begin with. Pencil in the directions of the electric fields caused by each charge in the places of interest. That will help guide you when it comes time to sum up their contributions at a given point (whether to add or subtract the magnitudes).

Calculate the contributions of each charge separately, one at a time, and then sum the results (this is allowed because electric fields obey the principle of superposition, which you should have learned about as a being a property of linear systems).

Regarding your Relevant Equations, (k)(q)/r gives the electric potential (in Volts) at distance r from a charge q. You're looking for the electric field strength, which has slightly different units: Volts per meter (or equivalently, Newtons per Coulomb). Change the 'r' to 'r2' in your formula.
Okay, but what would i do with the 3 distances mentioned in the last distance, 1m,2m,3m?
 
  • #5
gneill
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20,793
2,773
Okay, but what would i do with the 3 distances mentioned in the last distance, 1m,2m,3m?
Take them one at a time and locate them on the diagram that I hope you've drawn. For each of the locations sum the contributions to the electric field from the charges (taking into account their directions, which you penciled in --- right?).
 
  • #6
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Take them one at a time and locate them on the diagram that I hope you've drawn. For each of the locations sum the contributions to the electric field from the charges (taking into account their directions, which you penciled in --- right?).
Yes, i drawn the diagram. Your gonna add them up and the vector is going to be to the right since both charges are positive right?
 
  • #7
gneill
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Yes, i drawn the diagram. Your gonna add them up and the vector is going to be to the right since both charges are positive right?
That will depend upon where the points are located with respect to the charges. Place vectors on your diagram that indicate the directions of the fields produced by the charges at the points of interest.

The magnitudes of the forces from each charge are easily calculated from the charge and distance from the charge. The direction to assign depends upon the positions of the point and charge. The field "arrows" that you pencil in will tell you the direction.
 
  • #8
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That will depend upon where the points are located with respect to the charges. Place vectors on your diagram that indicate the directions of the fields produced by the charges at the points of interest.

The magnitudes of the forces from each charge are easily calculated from the charge and distance from the charge. The direction to assign depends upon the positions of the point and charge. The field "arrows" that you pencil in will tell you the direction.
Okay, but how would i set up the equation to the problem? I know the directions of the vectors are going to be facing each other. Qa is going to be toward B, and Qb is going to be towards A. But how would you set up the equation?
 
  • #9
gneill
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20,793
2,773
Okay, but how would i set up the equation to the problem? I know the directions of the vectors are going to be facing each other. Qa is going to be toward B, and Qb is going to be towards A. But how would you set up the equation?
Directions are with respect to the coordinate system that's established. From the given information you know where the X-axis origin is and the position of the charges on the X-axis. So you now know whether the contributions of the fields at any location on the X-axis should be positive or negative (negative if the arrow points to the left, positive if it point to the right).

The rest is just establishing the magnitudes of the field contributions using the formula. You know the positions of the charges and points, so you can establish the distances between the points and charges.
 

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