Electricity and magnetism. 2 questions

[Brewer]

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.

 

[Andrew Mason]

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, $\vec F = q\vec v \times \vec B = qvsin(x)B\hat k = .48N \hat k$

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

Plugging in the numbers:

$$q = .48/(.005 * 4000 * .602) = .039 C$$

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

AM

 

[Brewer]

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!