- #1
shimizua
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Homework Statement
A proton that is accelerated from rest through a potential of 20.0 kV enters the velocity filter, consisting of a parallel-plate capacitor and a magnetic field, shown below.
http://capa-new.colorado.edu/msuphysicslib/Graphics/Gtype61/prob09_velfilter.gif
The E-field between the parallel capacitor plates is 5.9x105 N/C. What B-field is required so that the protons are not deflected?
Homework Equations
he protons entering the region of the electric field E will experience a force directed downward with magnitude:
z*E, where z = +1 elementary charge
The protons will also experience a force due to the magnetic field of magnitude:
z*v*B,
where v is the speed of the protons. The direction of this force is given by the right hand rule, so for a magnetic field directed into the page, this force is directed toward the top of the page.
When these forces are equal in magnitude, then the proton will pass undeflected through the velocity filter. The condition of interest is then:
z*v*B = z*E
B = E/v
To find the speed of the protons, we need to calculate their kinetic energy:
(1/2)*m*v^2 = z*V,
where V is the accelerating voltage, so
v = sqrt(2*z*V/m)
B = E/sqrt(2*z*V/m)
E = 5.9*10^5 N/C
z = 1.602*10^-19 C
V = 20*10^3 V
m = ?
The Attempt at a Solution
I am just really confused on how to find what m is since i don't know what v is? like i know the big V but not the little v.