Electric field intensity/ charge magnitude question

  • #1

Homework Statement


Two equal but oppositely charged points are 1.0 m apart in a vacuum. The electric field intensity at the midpoint between the charges is 2.4x10^5 N/C. What is the magnitude of each charge?


Homework Equations


Electric force= Kc (q/r^2)
Kc=8.99x10^9


The Attempt at a Solution


Hi guys, first off, thanks for any help given.
My problem is that I am unsure as to how to incorporate both missing charges into the equation.
I tried solving the equation simply by putting in what is given.

2.4 x 10^5 = 8.99x10^9(q/1^2)
2.4x10^5=8.99x10^9 x q
q= .000027

Thus, I only found one charge, and I don't even believe that is right.. Could someone help explain what I am to do? Thanks for the help!
 

Answers and Replies

  • #2
hage567
Homework Helper
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You have to consider the sum of the electric field from each charge at the midpoint. Also, be careful with your distances. The distance from each charge to the point between them is 0.5 meter, not 1 meter.
 
  • #3
thanks, hage, but i'm still confused how the two charges are linked, how they fit together to create that 2.4 x 10^5 charge at their midpoint. i tried the problem again, solving it as so...
EF = Kc (q/r^2)
2.4 x 10^5 = 8.99x10^9 q/.25
60000= 8.99x 10^9
q = 6.67x10^-6
q= -6.67x10^-6

i feel like this is still incorrect... thanks again for the future help.
 
  • #4
hage567
Homework Helper
1,509
2
You must add up the electric field from the postive charge with the electric field from the negative charge at the midpoint. Remember, E is a vector quantity. The sum of the two gives 2.4x10^5 N/C at the midpoint. So the idea is E = E1 + E2.
 

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