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Determin the Electric Potential Difference

  • Thread starter kvarner83
  • Start date
2
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1. Homework Statement
The drawing shows a uniform electric field that points in the negative y direction; the magnitude of the field is 3000 N/C. Determine the electric potential difference (a) VB - VA between points A and B, (b) VC - VB between points B and C, and (c) VA - VC between points C and A.

http://edugen.wiley.com/edugen/courses/crs2216/art/qb/qu/c19/ch19p_56.gif

2. Homework Equations



3. The Attempt at a Solution
I was able to solve the first two parts of this problem easily but I am hung up on how the angle of the triangle plays into this.
I know A is zero because they are on the same equipotential field.
B I solved by converting the distance to meters and multiplying that by the magnitude of the electric field.
For C I know that it will be a combination of both the vertical and horizontal direction and I know that I am supposed to find that by using sin or cos but I just cant seem to put it together!
 

Delphi51

Homework Helper
3,407
10
Looks good so far, Kvarner.
At the risk of boring you, I suggest going back to the definition of electric potential as the work per charge needed to move a charge through an electric field. When moving the charge horizontally from A to B, there is no force in the direction of movement so no work need be done. Like moving horizontally in a gravitational field. So, moving from A to C you could do A to B first with work, energy and potential difference zero. Then add the potential difference for the B to C move that you've already calculated. Hope this makes sense ...

Alternatively, going straight from A to B the electric force is purely vertical so the pd is the work/charge = F*d/q = q*E*d/q where only the vertical component of distance is used because that is the direction of the force. I suppose it should really be written as a dot product F•d.
 

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