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Can somebody show me how to set up this problem?

A uniform electric field of magnitude 270 V/m is directed in the positive x direction. A +14^(-6) C charge moves from the origin to the point (x,y) = (20.0 cm, 50.0 cm)

A) What is the change in the potential energy of the charge field system?

B) Through what potential difference does the charge move?

I've solved part B (-54 volts), but I'm stumped as to how to tie it in with part A. Can somebody help?

2007-08-31 06:36:09 · 2 answers · asked by Anonymous in Science & Mathematics Physics

2 answers

What makes this easy is the statement, "uniform electric field...directed in the positive x direction". That means that moving around in y does absolutely nothing, and determining the field change is directly related to the change in x. This is about as simple a field as you'll ever see!

So, if it's 270 V/m, and you move 0.2 m, then 270*0.2=54. If you want to go the whole 9 yards, you could show that the field potential at (0,0) is 0, and the field potential at (0.2, 0.5) is 54, etc.

2007-08-31 06:46:04 · answer #1 · answered by El Jefe 7 · 0 0

Change in the potential energy is equal to to work done to move the charge between the two points.
You could calculate the force and multiply it with the distance (only components in X axis direction should be considered in this example, since no work is needed to move charge perpendicular to the field direction)


The easier way is to think of potential difference as difference in potential energy per unit charge. In this case we know the charge and the potential difference; all you need to do is to multiply the two values to get the energy difference.
-54V x 14 microC = -7.56 x 10^(-3) J
The potential energy decreased by 7.56 mJ.

2007-08-31 07:06:21 · answer #2 · answered by nosf37 4 · 0 0

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