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Charges on the vertex of a triangle 
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#1
May1607, 07:10 PM

P: 214

Positive pointcharges of +16.0 mC are fixed at two of the vertices of an equilateral triangle with sides of 1.30 m, located in vacuum. Determine the magnitude of the Efield at the third vertex.
so...all the particals are in equalibrum so that means F=ma=0 so if i call the third vertex c...Ea+Eb+Ec=0 so that means that Ea+Eb=Ec now i can write kqa/r^2+kqb/r^2=Ec and i just have to solve for Ec right? 


#2
May1607, 07:34 PM

HW Helper
P: 857

note that Efields are vectorial quantities. Ie. they have a direction. in your case, two point charges at two different location each exerting a Force radially, will induce two different field vectors at the the third spot. These two vectors are not parallel and so need to take the angle in between into account to get the right answer.
btw, the question doesn't seem to mention anything about the system being in equilibrium, just asking you work out the resulant field at third spot 


#3
May1607, 07:44 PM

P: 214

ummm...so do i have to find in they x and y compnents and all that sutff?



#4
May1607, 10:21 PM

P: 137

Charges on the vertex of a triangle
Don't panic  it's simple trig  sketch the two vectors at the 3rd vertex point keeping in mind that the E field for each charge is defined in terms of the force per unit positive test charge; i.e. the force which would be produced on a positive unit charge placed at the point under consideration. For your case, each charge would repel a unit positive test charge  notice the direction each would be trying to repel it. They're NOT the same direction. So, as you can see (if you really sketched the situation), you must find the vector sum of the two E vectors which, in this case, ends up being just the sum of their y components. BTW  this is a static problem  there's no motion involved. It's like an ideal, instantaneous snapshot just before any motion could take place. jf 


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