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!

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]
    [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?

  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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook