Mar 10, 2010 · 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.

Feb 3, 2008 · 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 …

Oct 30, 2007 · 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 …

Sep 1, 2010 · 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 = …

Oct 4, 2016 · 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 …

Sep 26, 2016 · 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 …

Nov 12, 2005 · 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 …

Sep 17, 2008 · 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 …

Mar 11, 2008 · 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 …

Mar 12, 2022 · We know the net force on the charged particle in the uniform electric field pointing up is mg - qE. To get acceleration, divide the net force by mass to get g - qE/m Plug into …