Calculate the change in electric potential energy

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1. Calculate the change in electric potential energy for a -10µC charge in a uniform electric field of strength 1750 N/C, traveling in a the straight paths OA, OB, or OC as shown. The figure shown is a rectangle with OA being a base, OC being a side and OB being a diagonal (No angle is given).



2. Homework Equations : Well i know ∆EPE = EPEf - EPEi. Also Electric Potential V = kq/r



3. Alright I need to find the change in Electric Potential Energy so I think this means finding the difference of the final and initial potentials. However no distance is given in the problem so V = kq/r won't help me. I'm thinking that since the electric field is 1750 N/C throughout the figure that maybe the EPE is constant which would give a change of 0. But I don't think the professor would bother giving me 3 questions all with 0 as an answer.
 

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  • #2
collinsmark
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If the electric field is uniform and non-zero, the potential is definitely not zero.

Here,
[tex] V = \frac{kq}{r} [/tex] won't help you, but not because no lengths are given, but rather because that relationship is only valid for point charges with respect to infinity. You're working with a uniform electric field here, not point charges.

Recall,

[tex] V = -\int _P \vec E \cdot d \vec l [/tex]

Since E is a constant, we can pull it out of the integral, integrate the path, making

[tex] V = -\vec E \cdot \vec l [/tex]

where [tex] \vec l [/tex] is merely your path vector.

If no specific lengths were given, I would suggest keeping the answers generalized. In other words, express your answers as a function of OA, OB, and OC.
 
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So is the path vector just the displacement of the path? Such as OA, OB and OC Meaning ∆V = -1750*OA? Also what affect does would the -10µC charge have on the potential if any?
 
  • #4
collinsmark
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So is the path vector just the displacement of the path? Such as OA, OB and OC Meaning ∆V = -1750*OA? Also what affect does would the -10µC charge have on the potential if any?

Keep in mind that electrical potential is not the same thing as potential energy. Rather, electrical potential is the potential energy per unit charge. So you'll have to multiply the change in electrical potential by the charge (in this case -10 µC) to get the change in potential energy.

Also, remember that you are dealing with vectors. I can't tell from the problem statement which direction [tex] \vec E [/tex] points. You may wish to work work with individual components of the vectors separately. For example, the potential change is zero if the path is perpendicular to [tex] \vec E [/tex]. This can have an effect on your final answers if you express your final answers are in terms of scalar lengths.
 
  • #5
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Ok this is starting to make some sense now. So the electric potential energy would be -(1750N/C)*OA and then if i multiply it by (1.0x10^-6C) im left with (1.75x10^-3)N*OA. After it's multiplied by OA it will be in joules which seems about right for potential energy.

Now, about the last thing you brought up about the direction of the vectors..
The diagram is layed out like this

C B


O A

With [tex]\vec{E}[/tex] starting at the right and traveling left. So this would make OC perpendicular which means the change is 0. OA is parallel so its just as i described above. And no angle is given for OB so maybe I should just leave it in terms of [tex]\theta[/tex].
 
  • #6
collinsmark
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Ok this is starting to make some sense now. So the electric potential energy would be -(1750N/C)*OA and then if i multiply it by (1.0x10^-6C) im left with (1.75x10^-3)N*OA. After it's multiplied by OA it will be in joules which seems about right for potential energy.

Actually, you might be off by a factor of 10. Earlier you mentioned the charge is -10 µC.

So I would say the change in potential energy is

[tex] PE(A) - PE(O) = -17.5 \times 10^{-3} J/m \times \left| AO \right| [/tex]

Now, about the last thing you brought up about the direction of the vectors..
The diagram is layed out like this

C B


O A

With [tex]\vec{E}[/tex] starting at the right and traveling left. So this would make OC perpendicular which means the change is 0. OA is parallel so its just as i described above.

Okay, so far so good. Just be sure to be careful with your signs.

And no angle is given for OB so maybe I should just leave it in terms of [tex]\theta[/tex].

Okay, here's where the nitty gritty of the vector stuff rears its head. Let me give you a hint.

[tex] V(B) - V(0) = - \vec E \cdot \vec {(OB)} = -E (OB) cos \theta = -E (OA) = V(A) - V(O) [/tex]

Which should make sense. You know that the potential difference between points B and A is zero (perpendicular to electric field). So the potential between points O and B must equal the potential difference between points O and A.

The electric potential (voltage, if you like) between two points is path independent. This comes from the fundamental theorem of calculus. The height at the top of a staircase minus the height at the bottom of the staircase is independent of how you actually get there. You could measure the height of each step and sum them together. Or you could go outside, climb the outside wall, go in through a window, all the while keeping track of your vertical distance, and you'd get the same answer.
 
  • #7
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Ok, I think i've got it. Thanks a lot for all of your help :biggrin:
 

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