Energy of Capacitor and Distance between Plates

[GomerPyle]

Alright guys, this stems from EK 1001 physics #846.

If the distance between the plates on the capacitor were doubled, the energy stored in the capacitor would:

A. Decrease by factor of 2
B. remain the same
C. Increase by factor of 2
D. Increase by a factor of 4


The way I want to solve this - V = ED. Increasing distance by factor of 2 between plates, increases the voltage between them by 2. Since C = Q/V, this decreases Capacitance by factor of 2.

The energy stored in capacitor is U = 0.5*Q*V. Since V is doubled, therefore U is doubled. This is not the right answer.........??

EK answer also states that "The voltage accross the capacitor is a function of the circuit and not the construction of the capacitor. Thus the voltage remains constant as long as only the capacitor changes and not the rest of the series." WHAT?? I thought when you increase the distance between the plates, this should alter the voltage between them via V=ED. This is what I also learned in my physics lab. Confused as hell, if anybody could chime in I would appreciate it. Thanks

Btw, answer is A.

---

Members don't see this ad.

 

[Saint Fu]

When you alter the geometry of a capacitor, what you're changing is the capacitance, C, not V. C is proportional to the area of the plates and inversely proportional to the distance between them. Thus, if the distance doubles, the capacitance is halved.

Like EK states, the voltage isn't dependent on the capacitor alone. Imagine a circuit that consists of a battery and a capacitor. It doesn't matter what the capacitance is, it will always store the same amount of voltage (i.e. the voltage of the battery). To reconcile this with V=Ed, consider that, as capacitance decreases, the amount of charge stored also decreases. As the amount of charge stored decreases, the strength of E decreases by a proportional amount. Thus, the doubling of d and the halving of E cancel out.

 

[cowboys]

Alright guys, this stems from EK 1001 physics #846.

If the distance between the plates on the capacitor were doubled, the energy stored in the capacitor would:

A. Decrease by factor of 2
B. remain the same
C. Increase by factor of 2
D. Increase by a factor of 4


The way I want to solve this - V = ED. Increasing distance by factor of 2 between plates, increases the voltage between them by 2. Since C = Q/V, this decreases Capacitance by factor of 2.

The energy stored in capacitor is U = 0.5*Q*V. Since V is doubled, therefore U is doubled. This is not the right answer.........??

EK answer also states that "The voltage accross the capacitor is a function of the circuit and not the construction of the capacitor. Thus the voltage remains constant as long as only the capacitor changes and not the rest of the series." WHAT?? I thought when you increase the distance between the plates, this should alter the voltage between them via V=ED. This is what I also learned in my physics lab. Confused as hell, if anybody could chime in I would appreciate it. Thanks

Btw, answer is A.

This has tripped me up so much. But I think what they are saying is right, the voltage (potential difference) will not change, since the battery doesn't change. Since the distance is doubled the capacitance, C = (k*e_0*A)/d, will decrease by a factor of 2. Since U = .5*C*V^2, energy stored will also decrease.

I hope that helps. If you're still not sure, crack open your physics text book. I'm sure it's collecting dust like mine

When you alter the geometry of a capacitor, what you're changing is the capacitance, C, not V. C is proportional to the area of the plates and inversely proportional to the distance between them. Thus, if the distance doubles, the capacitance is halved.

Like EK states, the voltage isn't dependent on the capacitor alone. Imagine a circuit that consists of a battery and a capacitor. It doesn't matter what the capacitance is, it will always store the same amount of voltage (i.e. the voltage of the battery). To reconcile this with V=Ed, consider that, as capacitance decreases, the amount of charge stored also decreases. As the amount of charge stored decreases, the strength of E decreases by a proportional amount. Thus, the doubling of d and the halving of E cancel out.

Nice explanation!

 

[GomerPyle]

Thanks guys - makes good sense. I just dont know why in my physics lab that I learned that increasing the distance between capacitor plates increases the voltage between them. Maybe my TA was an idiot...but I remember it being the solution to one of the problems.....ugh oh well.

 

[Saint Fu]

Thanks guys - makes good sense. I just dont know why in my physics lab that I learned that increasing the distance between capacitor plates increases the voltage between them. Maybe my TA was an idiot...but I remember it being the solution to one of the problems.....ugh oh well.

Say that the capacitor was charged and then disconnected from the circuit. In this case, V between the plates will vary with the distance between them because you're holding the charge on each plate constant. Assuming the space between the plates is very small compared to their dimensions, I think that we can assume that E remains constant as d varies, because E is uniform (though you and I might want to think about that a little more to make sure).

Basically, in the case of a circuit, the voltage of the battery and the capacitance determine the amount of charge collected on the plates . In the case of an isolated pair of parallel plates, the amount of charge on the plates and the distance between them determine the voltage between them. In each case, you're holding different values constant.

 

[GomerPyle]

Say that the capacitor was charged and then disconnected from the circuit. In this case, V between the plates will vary with the distance between them because you're holding the charge on each plate constant. Assuming the space between the plates is very small compared to their dimensions, I think that we can assume that E remains constant as d varies, because E is uniform (though you and I might want to think about that a little more to make sure).

Basically, in the case of a circuit, the voltage of the battery and the capacitance determine the amount of charge collected on the plates . In the case of an isolated pair of parallel plates, the amount of charge on the plates and the distance between them determine the voltage between them. In each case, you're holding different values constant.

That last paragraph really helped me understand this. Thank you. However, I do not understand which values I hold constant in each case, and why? Do you have time to explain that?

 

[Saint Fu]

That last paragraph really helped me understand this. Thank you. However, I do not understand which values I hold constant in each case, and why? Do you have time to explain that?

I guess you should think about the system in each case. In the simple capacitor/battery example, Kirchoff's voltage/loop rule states that the voltage across the capacitor has to equal the voltage supplied by the battery. In this case, the voltage is applied externally to the capacitor, so it's an independent variable, while things like charge/energy stored is the dependent variable.

When you have a pair of charged plates, I think it's intuitive to assume that the charge on the plates is constant, unless it's stated that there's a discharge at some point. In the end it depends on the problem/passage. Capacitors have transient characteristics, and I'm sure the MCAT writers can come up with whatever devious scenarios so there's not really a one-size-fits-all answer, but hopefully this helps with intuition.