1. Not finding help here? Sign up for a free 30min 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!

Electricity and magnetism. 2 questions

  1. Mar 7, 2006 #1
    A couple of questions.

    1. A capacitor of capacitance C is discharging through a resistor of resistance R. When will the energy stored be equal to half its initial value?

    My logic went like this:

    Energy in a capacitor (U) = Q^2/2C
    so: Q = (2CU)^0.5

    so therefore the charge when the energy is halved = (2C(U/2))^0.5 = (CU)^0.5

    so from Q=Q0*exp(-t/RC)
    (CU)^0.5 = (2CU)^0.5*exp(-t/RC)
    (CU)^0.5 = (2)^0.5*(CU)^0.5*exp(-t/RC)
    t = -RCln(1/(2)^0.5)

    and this second question I don't really follow at all.

    A particle of mass 6.0g moves at 4.0km/s in an xy plane, in a region of space with uniform magnetic field given by 5.0imT. At one instant when the particles velocity is directed at 37degrees counter clockwise from the positive x direction, the magnetic force on the particle is 0.48kN. What is the particles charge?

    I had a go using: F=qvBsin(x) (where x = angle) and got a charge of 0.039C. Is this the right way to go about it? Seems quite a large charge to me.
  2. jcsd
  3. Mar 8, 2006 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Your answer to the first question is correct.

    In the second, I think you are also right. I am assuming that i is the unit vector in the x direction and k is the unit vector in the z direction.

    So, [itex]\vec F = q\vec v \times \vec B = qvsin(x)B\hat k = .48N \hat k[/itex]

    [tex]\frac{F}{Bvsin(x)} = q [/itex]

    Plugging in the numbers:

    [tex]q = .48/(.005 * 4000 * .602) = .039 C[/tex]

    Think of the moving charge in terms of current: dQ/dt = I where dQ is the charge strung out over 4 km. The current is .039 amps. So this is equivalent to the force on a 4 km section of wire conducting a current of .039 amps

  4. Mar 8, 2006 #3
    So I wasn't far off? Thats not bad. I was just concerned because a mass was given in the question,but I didn't seem to use one.

    Ok. Thanks for that. Thats a bit more confidence in myself there!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electricity and magnetism. 2 questions
  1. Magnetism question 2 (Replies: 1)