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How does electric current change electric potential?

  1. Feb 16, 2010 #1
    Suppose we have an electrically insulated box in a room, and the potential difference between the box and the room is given by [tex]V_1=V_{Box1}-V_{room}[/tex].

    Now suppose we inject the box with X amps of current, for t seconds. The new voltage is now given by [tex]V_2=V_{Box2}-V_{room}[/tex]

    By how much does the injection of X amps of current change the electric potential of the box?
    What is [tex]V_{Box1}-V_{Box2}[/tex]?
    Last edited: Feb 16, 2010
  2. jcsd
  3. Feb 16, 2010 #2
    It depends on the capacitance of the box. V=Q/C
    and Q = It
    Your current has placed more charge on the box so its potential will rise according to the value of its capacitance.
    Use V=Q/C and write an expression for V1 and V2 in terms of the charge on the box and its capacitance.
    Last edited: Feb 16, 2010
  4. Feb 16, 2010 #3
    So can we equate the two equations you've written and simply say [tex]CV=It[/tex]?

    if the box is a good electrical insulator, then we expect it to have a high capacitance (ability to hold an electrical charge), right?
    But then by the CV=It above, the higher the C is the less effect a current I will have on the potential V. What's the intuitive meaning behind this?
  5. Feb 16, 2010 #4
    The initial pd on the box is given by V1=Q/C where Q is the initial charge and C its capacitance.
    The final pd of the box (V2) is found by putting the charge on it equal to Q + It (and the same value of C.)
    You then find V2 - V1
    The definition of C is that it equals Q/V
    The intuitive part of this is that the higher the value of C, the more charge you need to increase the pd. So as current is flow of charge, the higher the capacitance means that you need more current to increase the pd.
    I always think of water in a container. The pd is the depth of the water, and the capacitance is the capacity (volume) of the container. The bigger the capacity of the container, the more water you need to raise its level.
    If you were filling it with a hose pipe, you would need a larger water current to fill a larger container to the same depth in the same time.
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