Electric Potential voltage and conservation of energy

In summary: The same applies to charged particles.In summary, the problem involves finding the final kinetic energy of a proton accelerated by a uniform electric field and experiencing a potential decrease of 100 V. The change in potential energy is equal to the work done on the proton, which can be calculated using the definition of potential difference. The answer obtained for the change in kinetic energy must be positive, as the proton has a negative potential energy and therefore a positive kinetic energy. This aligns with the law of conservation of energy, and any negative values for kinetic energy are incorrect.
  • #1
therealkellys
16
0
Hi,

I'm a little stuck on a problem involving voltage and conservation of energy. The answer that I came up with involved a negative value for kinetic energy, which doesn't make sense to me because mass and v^2 are never negative.

Problem: A proton is accelerated from rest by a uniform electric field. The proton experiences a potential decrease of 100 V. Find its final kinetic energy.

What I've done so far:
KE_i = 0 because accelerated from rest
V_b - V_a = -100 V
Change in voltage = Work/charge = [-q(int)E dot ds] / q = Change in potential energy

-100 = -Change in PE --> 100 J = change in potential energy

Therefore, under law of conservation of energy... change in kinetic energy must equal -100 J. Since KE_i = 0 J, KE_f= -100 J. :confused:

Thanks in advance for the help.
 
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  • #2
Bump... I'm still really unsure about my answer. If anybody could point out anything that I might have overlooked, that would help a lot. Thanks!
 
  • #3
therealkellys said:
The answer that I came up with involved a negative value for kinetic energy, which doesn't make sense to me because mass and v^2 are never negative.
Your suspicions are correct. Change is usually expressed as final value minus the initial value, hence leading to a positive value for the change in potential energy. Just for reference a perhaps easier method would be to consider the definition of potential difference or voltage; which is work done per unit charge, thus;

[tex]V = \frac{W}{q}[/tex]

From which we can obtain the expression [itex]W = Vq[/itex], substituting your values in we obtain [itex]W = 100 \times 1 = 100\; J[/itex].
 
Last edited:
  • #4
[tex]q \Delta V = -\Delta E_{kin}[/tex]
 
  • #5
Since your proton is at a lower potential, it has a negative potential energy (-100J), so its kinectic energy must be +100J.
Think of it in terms of mechanics. You have a mass of 1kg at a height of 10m. You have a potential energy of 100J (actually 98) in reference to ground. If you drop the mass, when it hits the ground its potential energy will be zero and the kinectic energy 98J.
 

1. What is electric potential?

Electric potential is the amount of electric potential energy per unit charge at a specific point in an electric field. It is a measure of the electric potential energy that a charged particle would have at that point.

2. How is electric potential related to voltage?

Voltage is the measure of the difference in electric potential between two points in an electric field. It is directly proportional to the electric potential difference between those points, with the constant of proportionality being the charge of the particle.

3. How does conservation of energy apply to electric potential and voltage?

Conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the case of electric potential and voltage, energy is conserved as it is converted from potential energy to kinetic energy when a charged particle moves in an electric field.

4. What is the unit of measurement for electric potential and voltage?

The unit of measurement for electric potential is the volt (V), while voltage is also measured in volts. One volt is equal to one joule per coulomb (J/C).

5. How does distance affect electric potential and voltage?

Distance plays a significant role in determining the electric potential and voltage in an electric field. As distance increases, the electric potential decreases, resulting in a lower voltage. This is because the electric field becomes weaker with distance, leading to a decrease in the amount of potential energy per unit charge.

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