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Electric field and electric force

  • Thread starter -EquinoX-
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  • #1
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Homework Statement



What is the electric field at location of charge q

2d2dudh.jpg


Homework Equations





The Attempt at a Solution



My intuition in solving this is by finding the field of each charges relative to q and add them all, so for example for the 2q charges relative to q, the electric forces it contributes is ke * 2q/a^2, am I right?
 

Answers and Replies

  • #2
rl.bhat
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You have to find the magnitude and direction of forces due to three charges at q, and then find the vector sum of all forces.
 
  • #3
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That's what I was trying to say, so say the magnitude of the force 2q at q is ke * 2q/a^2 in the positive x direction (or i hat) , am I right?
 
  • #4
rl.bhat
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Yes. You right. Similarly find the forces due to 3q and 4q, and find the vector sum.
 
  • #5
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hmm..for the 3q charge, the distance to q is sqrt(2)a, so therefore the magnitude can be found by:

ke * 3q/(sqrt(2)a)^2 cos(45) i + ke * 3q/(sqrt(2)a)^2 sin(45) j

is this correct?
 
  • #6
rl.bhat
Homework Helper
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hmm..for the 3q charge, the distance to q is sqrt(2)a, so therefore the magnitude can be found by:

ke * 3q/(sqrt(2)a)^2 cos(45) i + ke * 3q/(sqrt(2)a)^2 sin(45) j

is this correct?
You can directly wright down as ke*3q/2a^2 for 3q and ke*4q/a^2.
Field due 2q and 4q are perpendicular to each other. Find the resultant of these fields. Add it to the field due to 3q to get the final field.
 
  • #7
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>>Field due 2q and 4q are perpendicular to each other. Find the resultant of these fields.

Well I already did find the field due 2q and q right? Why do I have to find between 2q and 41? The question asks with respect to q
 
  • #8
rl.bhat
Homework Helper
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The field at location of charge q is the vector sum of fields due to 2q, 3q and 4q.
 
  • #9
564
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Thats trueand so why is my answer 2 posts above is wrong?
 

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