Which PE equation do I use?

[going2breakdown]

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P=(1/2)QV and P=qV When do I use one or the other. This concerns E and M.

 

[Zoidburg]

P=(1/2)QV and P=qV When do I use one or the other. This concerns E and M.

In short, P=qV is the general formula for finding electrical potential energy and P=(1/2)QV is a special case of that formula for capacitors.

If that doesn't satisfy you, I'll go on to more detail below which will hopefully help.

PE=qV
This comes from the definition of voltage which is energy contained per coulomb of charge so:

V=energy/q

Thinking of this energy as potential energy and rearranging the equation we end up with what you mentioned(P=qV). As for uses of this equation, one classic example can be found here:

http://answers.yahoo.com/question/index?qid=20090121192721AAftcJg

In this, V=Ed is used first to find V, then P=qV is used. V=Ed is another handy formula, just remember that the d refers to the distance traveled parallel to the electric field and perpendicular distance does not count because energy increases or decreases only in the direction of the force (think of W=Fd) so you may need to use some trigonometry in some problems to get the proper distance.

U=(1/2)QV
This is the above formula applied to capacitors. Since capacitors go from 0 Volts to some maximum voltage as they charge, we use (1/2)QV to take an average of the energy involved. You should also recognize that the formula for capacitance, C=Q/V can be used to make different forms of (1/2)QV through substitution so:

Energy stored in a capacitor(U)=1/2 QV = 1/2 CV^2 = (1/2 Q^2)/C

One other thing to know involving this formula is its relation to C=kA/d where C is capacitance, k is the dielectric constant, A is area of the capacitor plate, and d is distance between the plates. It's good to use these 2 formulas to better understand what happens when changing aspects of a capacitor. For instance, you can see that increasing the distance between plates would lower capacitance, but increase energy stored(U). (Always use U= (1/2 Q^2)/C during this type of logic check because V is constantly changing as the capacitor charges so U=1/2CV^2 is useless for this because you want all other variables besides C constant. Q is constant which is why the other form is used.

 

[V5RED]

In short, P=qV is the general formula for finding electrical potential energy and P=(1/2)QV is a special case of that formula for capacitors.

If that doesn't satisfy you, I'll go on to more detail below which will hopefully help.

PE=qV
This comes from the definition of voltage which is energy contained per coulomb of charge so:

V=energy/q

Thinking of this energy as potential energy and rearranging the equation we end up with what you mentioned(P=qV). As for uses of this equation, one classic example can be found here:

http://answers.yahoo.com/question/index?qid=20090121192721AAftcJg

In this, V=Ed is used first to find V, then P=qV is used. V=Ed is another handy formula, just remember that the d refers to the distance traveled parallel to the electric field and perpendicular distance does not count because energy increases or decreases only in the direction of the force (think of W=Fd) so you may need to use some trigonometry in some problems to get the proper distance.

U=(1/2)QV
This is the above formula applied to capacitors. Since capacitors go from 0 Volts to some maximum voltage as they charge, we use (1/2)QV to take an average of the energy involved. You should also recognize that the formula for capacitance, C=Q/V can be used to make different forms of (1/2)QV through substitution so:

Energy stored in a capacitor(U)=1/2 QV = 1/2 CV^2 = (1/2 Q^2)/C

One other thing to know involving this formula is its relation to C=kA/d where C is capacitance, k is the dielectric constant, A is area of the capacitor plate, and d is distance between the plates. It's good to use these 2 formulas to better understand what happens when changing aspects of a capacitor. For instance, you can see that increasing the distance between plates would lower capacitance, but increase energy stored(U). (Always use U= (1/2 Q^2)/C during this type of logic check because V is constantly changing as the capacitor charges so U=1/2CV^2 is useless for this because you want all other variables besides C constant. Q is constant which is why the other form is used.

The general formula for electric potential energy is P=k*q1*q2/r

P=qV and P=1/2qV are both special cases involving electric potential energy.

P=qV applies to uniform, constant electric fields, and it is the amount of energy it takes to move a charged particle from one side of the uniform, constant, electric field to the other(or to prevent it from moving). This can be in a capacitor. If a capacitor has been charged, then there exists an electric field between the plates, so a particle in that field can have a potential energy due to its position in the field. The q in this formula is the charge on the particle, and the V is the electric potential due to the field created by the capacitor.

P=1/2qV applies to a capacitor, and it is the amount of energy currently stored in a capacitor that has been charged. The q in this formula is the magnitude of charge on either plate of the capacitor, and the V is the electric potential due to the field created by the capacitor. I have no idea where you got the idea that it is the average energy stored. I have not found that idea anywhere but in your post. I checked my textbook as well as many websites and they all say that P=1/2qV is the energy currently stored in a capacitor, not the average of its maximum and its minimum.

 

[Zoidburg]

The general formula for electric potential energy is P=k*q1*q2/r

P=qV and P=1/2qV are both special cases involving electric potential energy.

P=qV applies to uniform, constant electric fields, and it is the amount of energy it takes to move a charged particle from one side of the uniform, constant, electric field to the other(or to prevent it from moving). This can be in a capacitor. If a capacitor has been charged, then there exists an electric field between the plates, so a particle in that field can have a potential energy due to its position in the field. The q in this formula is the charge on the particle, and the V is the electric potential due to the field created by the capacitor.

P=1/2qV applies to a capacitor, and it is the amount of energy currently stored in a capacitor that has been charged. The q in this formula is the magnitude of charge on either plate of the capacitor, and the V is the electric potential due to the field created by the capacitor. I have no idea where you got the idea that it is the average energy stored. I have not found that idea anywhere but in your post. I checked my textbook as well as many websites and they all say that P=1/2qV is the energy currently stored in a capacitor, not the average of its maximum and its minimum.

Yes, PE=k*q1*q2/r is the more common form one should recognize, but you can convert between it and PE=qV using V=kq/r. One thing to note is how similar PE=k*q1*q2/r is to Coulomb's law F=kq1*q2/r^2 so make sure the right one is used for the right situation.

About it being an average, that was a poor explanation on my part. I was referring to how it is derived by taking the average value or integral of V with respect to charge and plugging in q/C for V which equals 1/2q^2/C or 1/2qV. Even though Calculus is not a needed skill for the MCAT, if anyone is interested, this link does a good job of explaining this derivation as well as other concepts the OP was asking about:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c2

 

[sciencebooks]

The general formula for electric potential energy is P=k*q1*q2/r

P=qV and P=1/2qV are both special cases involving electric potential energy.

P=qV applies to uniform, constant electric fields, and it is the amount of energy it takes to move a charged particle from one side of the uniform, constant, electric field to the other(or to prevent it from moving). This can be in a capacitor. If a capacitor has been charged, then there exists an electric field between the plates, so a particle in that field can have a potential energy due to its position in the field. The q in this formula is the charge on the particle, and the V is the electric potential due to the field created by the capacitor.

P=1/2qV applies to a capacitor, and it is the amount of energy currently stored in a capacitor that has been charged. The q in this formula is the magnitude of charge on either plate of the capacitor, and the V is the electric potential due to the field created by the capacitor. I have no idea where you got the idea that it is the average energy stored. I have not found that idea anywhere but in your post. I checked my textbook as well as many websites and they all say that P=1/2qV is the energy currently stored in a capacitor, not the average of its maximum and its minimum.

Thanks for this.

 

[going2breakdown]

Thank you everyone.