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Potential Difference and Capacitors 
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#1
Feb2113, 04:13 PM

P: 1

So I am having some trouble conceptualizing potential difference, and how to calculate it without integrating the EField. My problem is how exactly would you calculate the potential difference between the two plates of a capacitor. When a capacitor is charged up with a battery of voltage V, the voltage between the plates also increases to V. However, I have no idea why that is, and no idea how you could calculate the potential difference between them. I am also wondering how you could calculate the potential difference between two known point charges.
I am aware of Q = CV and all the other equations for capacitors, I just dont understand the reasoning behind V = Q/C etc. 


#2
Feb2213, 05:06 PM

P: 102

When you charge a capacitor, electrons are added to one plate and depleted on the other until the static potential from the charges balances the applied voltage. At that point, the current stops flowing. Simply put, the electrons want to go from the negative terminal to the positive terminal on the battery, but are blocked by the insulator between the two plates. As a result the electrons pile up in the plates which sets up an opposing efield.



#3
Feb2213, 06:05 PM

P: 389

That means Q = V*4*pi*e*r Therefore V = Q/(4*pi*e*r) So if you have two spheres and their distance is much bigger than their radius the potential difference between them is approximately 1/(4*pi*e) * (Q1/r1  Q2/r2) Since point charges have a radius of 0 you can not calculate their potential. 


#4
Feb2313, 01:46 PM

P: 77

Potential Difference and Capacitors
I understand that the electrons stop flowing as they are blocked by the insulator and the accumulated electrons. But how is the same effect achieved by an inductor, i.e., when the potential difference accross an inductor equals that of the battery, no current flows, but what is blocking them now?



#5
Feb2313, 02:01 PM

P: 102

Inductors block current due to changing the magnetic field. The field is created when current starts to flow thru the inductor which creates an opposing voltage (counteremf). More properly, inductors oppose changes in current because any changes in current produces a change in its magnetic field which in turn produces an opposing voltage which tries to keep the current steady. Capacitors store energy in the form of electric charge. Inductors store energy in magnetic fields.



#6
Feb2313, 02:03 PM

P: 77

The analogy im talking about is visualized in 11:25 in here http://www.youtube.com/watch?v=U_gj9B0llE 


#7
Feb2313, 03:47 PM

P: 102

The changing magnetic field in the inductor produces a voltage opposing the change in current which can be thought of as a battery in series with the circuit.



#8
Feb2313, 05:40 PM

P: 77

Ok, so after asking here: http://openstudy.com/study#/updates/...b0111cc6900335 and elsewhere, I think I got to this conclusion, which I'm not sure is right. I'm copypasting from a fb conversation I had:
" in electronics everyone works with the battery being always at the same voltage i mean the connections of the battery but i think it's not true if you have an inductor say it changes the voltage at those points but what people do is to say that a net voltage difference develops across the ends of an inductor, so that the voltage difference across resistors is the same as doing the right thing and assume that even though theres a net voltage across the inductor it doesnt change a thing" Is this right? 


#9
Feb2413, 12:46 AM

P: 102

Not sure what the statement about the batteries means. You can model the inductor as a battery, or more properly as a current dependent voltage source whose output is based on the equation:
V = L * di/dt where V is voltage, L is inductance, and di/dt is the change in current thru the inductor with time. Just remember that the sum of the voltages around a closed circuit loop must equal zero. 


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