2010年3月10日 · F=qE and F=q(V*B) The first is just a vector relation for a charge in an electric field. The second is the Lorentz force on a moving charged particle in a magnetic field.

2008年2月3日 · In the case of a velocity selector, the magnetic field is always at 90 degrees to the velocity and the force is simplified to F = qvB in the direction described be the cross product. F = qE and F = qvB So, qE = qvB V = E/B Setting the two forces to equal magnitude in opposite directions it can be shown that V = E/B .

2007年10月30日 · F = qE Acceleration (a) = F / m = qE/m mechanics: x = Vi*t y = - (qE/2m)t^2 Vfx = Vi Vfy = -(qE/m)*(1/Vi) I think I have most of what I need here. This next figure illustrates how I think the two trajectories take place, and I wanted to check this before I went further.

2010年9月1日 · so, the question is basically asking me to rearrange F = qE + qvB to find 'v'. The Attempt at a Solution I'm not the greatest at rearrangements, but here is my attempt. F - qvB = qE F - v = qE / qB (q values cancel each other out) v = (E / B) + F is this right, or have i gone wrong somewhere? Any help would be much appreciated :)

2005年11月12日 · How does one prove that the force between two conductor plates is F=q^2/2eA, where e=epsilon. If I use F=qE. Where the electric field is generated by one of the conducting plates: E=q/eA, then F=q^2/eA. I think what the answer assumes is that the E-field is from a uniform plate (insulator): E=q/2eA. In that case, F=q^2/2eA, which is the correct ...

2018年7月16日 · F⋅Δs = qE⋅Δs. F and E are parallel; equipotential lines surround a charge radially. Since no work is done to move a charge along an equipotential line, F⋅Δs = 0. This means qE⋅Δs = 0, but q, E, and Δs are assumed to be nonzero. The only way to make the equation zero is to make E perpendicular to Δs (i.e., cosθ = 0) in all cases.

2016年10月4日 · So, F/q=E(x). which would give us E(x)=k(q1* r(hat))/r^2 They both calculate the force, coulombs is the force between two charges and the electric field is the force a some point q.

2016年9月26日 · First of all, I found that there are two equations for the Lorentz force: one of them is F = qE + qv × B ... Insights Blog -- Browse All Articles -- Physics Articles Physics Tutorials Physics Guides Physics FAQ Math Articles Math Tutorials Math Guides Math FAQ Education Articles Education Guides Bio/Chem Articles Technology Guides Computer ...

2008年9月17日 · If an oil drop with mass 1.00x10^-14 kg loses an electron whilst in an electric field of 1.00x10^6 N/C, what is the change in acceleration? *** F=ma F_e = qE So I know delta a for change in acceleration will be (qE/m) final - (qE/m) initial... but …

2008年3月11日 · F=qE'+q(v+v')xB'=qE'+q(vxB')+q(v'xB')=mr''=qE+vxB it shouldn't be to difficult to find the problems from there. note: the E' and B' are the gallilean transform of the magnetic and electric fields, if you check out maxwell's equations you should find that its not possible for the gallilean transform to yield the same equations of motion.