Recent content by Travkid

Ah right! Of course. The Potential Difference. The equations I know for Electric Field are... E= Force/q0 E= kq/r But where does the potential difference come into play? The P.d is essentially a voltage, right?

What is a p.d? We also haven't talked about work on charged particles at all. I'm taking the general algebra based physics, not the calculus based one.

Homework Statement An electron accelerated from rest through a voltage of 420 V enters a region of constant magnetic field. If the electron follows a circular path with a radius of 29 cm, what is the magnitude of the magnetic field? Homework Equations r=mv/lqlB K=VQ Conservation of Energy...

Thank you. End up getting the answer :)

Uh oh... I think I'm lost then lol. Well now I have the potential energy... but I'm not sure how to use this to find the distance. I thought I could use the Formula I wrote above.. U =kq0q/r but it doesn't seem right. Am I going to use the eletric potential of a point charge at 0 and infinity? V...

So the PE is equal to... .078992001 - .019748 = .059244001 Joules. .059244001 = kQ0q/r I know what both of my Q0, k, q. But I have the energy in Joules and I know I need to convert it into coulombs... I'm working with a positive charge so I need to divide my Potential Energy by the charge of a...

KE at half the speed to infinity: 1/2(0.0029)(3.690435^2 = .019748 Joules But we know that when the speed is at infinity, potential is equal to 0, right? Therefore the Potential Energy must be: KE at speed at Infinity - KE at half the speed to infinity so. PE = 1/2(0.0029)(7.38087)^2 -...

7.38087 was the velocity. KE = 1/2mv2 So the total KE is 1/2(0.0029)(7.38087)^2 Total Energy must be equal to KE + PE... hmm. But to find the Potential Energy... (8.99 x 10^9)(3.37 x 10^-6)^2 / Sqrt(1.6707) So the Total Energy is both of those put together. But how is the total energy going...

Homework Statement [/B] A charge of +3.37µC is held fixed at the origin. A second charge of 3.37µC is released from rest at the position (1.15 m, 0.590 m) . (a) If the mass of the second charge is 2.90 g, what is its speed when it moves infinitely far from the origin?(b) At what distance from...