1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Potential Difference and Capacitors

  1. Feb 21, 2013 #1
    So I am having some trouble conceptualizing potential difference, and how to calculate it without integrating the E-Field. 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. jcsd
  3. Feb 22, 2013 #2
    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 e-field.
  4. Feb 22, 2013 #3
    The capacity of a sphere is 4*pi*e*r.
    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.
    Capacitance is defined as Q/V. Therefore V = Q/C by definition.
  5. Feb 23, 2013 #4
    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?
  6. Feb 23, 2013 #5
    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 (counter-emf). 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.
  7. Feb 23, 2013 #6
    Ok, so imagining the circuit as the typical analogy of stairs going up (battery) and down (resistor).. how would the inductor, look? If it produces an opposing voltage should it look like a battery then (i.e. sloping in the opposite direction than the resistor), or what?

    The analogy im talking about is visualized in 11:25 in here
    Last edited by a moderator: Sep 25, 2014
  8. Feb 23, 2013 #7
    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.
  9. Feb 23, 2013 #8
    Ok, so after asking here: http://openstudy.com/study#/updates/51291abfe4b0111cc6900335 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?
  10. Feb 24, 2013 #9
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook