Recent content by Koh Eng Kiat

Thanks! I solved it already!

Do you need to change A to 1/2A?

Homework Statement The plates of a parallel plate capacitor of capacitance C0 are horizontal. Into the gap a slab of dielectric material with K=2 is placed, filling the bottom half of the gap between the plates. What is the resulting new capacitance? Homework Equations C = Q/V = κɛ0A/d The...

Thanks!

r = mv/qb v = 3.142 / 0.63 x 10^-6s = 4986655 m/s 2 = (9.11 x 10^-31kg)(4986655m/s)/(1.6 x 10^-19C)(b) 2 = (4.54 x 10^-24)/(1.6 x 10^-19C)(b) b = (4.52 x 10^-24)/(3.2 x 10^-19C) = 14 micro T. The answer is correct.Thanks for the help. It was calculation mistake on my part. However, isn't...

a.18 μT b.14 μT c.28 μT d.34 μT e.227 μT The answer is b, according to my answer given. I get a much larger answer using the quarter circle's path length. Oh man...

So I should use the circumference of the segment of the circle? The distance would be (2 x pi x 2m) / (4) = pi. Using pi / time to find velocity? Sorry guys. I have been trying quite hard on this question. What would be the ideal solution then?

I now know that the trajectory of the electron is curved upwards. The reason is that the velocity direction is perpendicular to the direction of the uniform magnetic field. Hence, the centripedal force will result in a curved motion. Therefore, it is the top half of the diagram. For my question...

Guys, thanks for replying! But I do not understand. What is the path that the electron takes? Why is my velocity correct? Isnt it displacement over time? I googled and know that the radius of a square is from the center to any of the four vertices, which I calculated to be 1.41m.

Homework Statement The boundary shown is that of a uniform magnetic field directed in the positive z direction. An electron enters the magnetic field with a velocity pointing along the x-axis and exits 0.63 μs later at point A. What is the magnitude of the magnetic field? Homework Equations...