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Homework Help: Electric Potential, Capacitors

  1. Apr 1, 2004 #1
    I have a few problems I'm stuck on, I'm not sure what equations to use...please help

    1) An electric car accelerates for 6.4 s by drawing energy from its 300-V battery pack. During this time, 1800 C of charge pass through the battery pack. Find the minimum power rating of the car.

    2) Location A is 2.70 m to the right of a point charge q. Location B lies on the same line and is 5.10 m to the right of the charge. The potential difference VB - VA = 40 V. What is the magnitude and sign of the charge?

    3) The electric potential energy stored in the capacitor of a defibrillator is 97 J, and the capacitance is 160 uF. What is the potential difference across the capacitor plates?
     
  2. jcsd
  3. Apr 1, 2004 #2
    1) The power rating, P, is:
    [tex]P = IV[/tex]
    The current, I, is:
    [tex]I = \frac{dq}{dt}[/tex]
    Therefore:
    [tex]P = \frac{dq}{Vdt}[/tex]

    2) The potential created by the charge q at a distance x from it is:
    [tex]V = k\frac{q}{x}[/tex]
    You can find this formula in your textbook. So the potential difference between two points that are distanced x1 and x2 from the point charge is:
    [tex]\Delta V = k(\frac{q}{x_1} - \frac{q}{x_2})[/tex]
    Rearrange and solve for q:
    [tex]q = \frac{\Delta V}{k}\frac{1}{\frac{1}{x_1} - \frac{1}{x_1}}[/tex]

    3) The electric potential energy of a charged capacitor is:
    [tex]E = \frac{1}{2}cV^2[/tex]
    Where V is the potential difference across the plates. So:
    [tex]V = \sqrt{\frac{2E}{c}}[/tex]
     
    Last edited: Apr 1, 2004
  4. Apr 1, 2004 #3
    Ok, I'm still not real sure how to solve numbers 1 and 3 and for the second problem I got an answer of 6.268x10^10 which I know is incorrect. I used 8.99x10^9(40/2.70-40/5.10). Where did I go wrong?
     
  5. Apr 1, 2004 #4
    It is the potential difference that you know, not the charge of the particle. You used 40v as the charge, which is clearly wrong. I have edited my post for more clarification but this is as far as I can go.
     
  6. Apr 1, 2004 #5
    Alright I've figured out problems 2 and 3 but I'm still stuck on the first one. What does the d mean in your equation?
     
  7. Apr 1, 2004 #6
    [tex]I = \frac{dq}{dt}[/tex]

    means that the current is equal to the derivative of the charge with respect to time.

    Have you had a course in calculus?

    cookiemonster
     
  8. Apr 2, 2004 #7
    The current, I, is defined as the amount of charge that goes through any given cross section in a given time period. So if 2c passed in 2 seconds, the current is 1A.
     
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