The E-field between the parallel capacitor plates is 4.5x10 5 N/C . What B-field

Question

The E-field between the parallel capacitor plates is 4.5x105N/C. What B-field is required so that the protons are not deflected? (Ignore relativistic effects for high velocities.)

A proton, that is accelerated from rest through a potential of 18.0kV enters the velocity filter, consisting of a parallel-plate capacitor and a magnetic field, shown below. The E-field between the parallel capacitor plates is 4.5x10^5N/C. What B-field is required so that the protons are not deflected? (Ignore relativistic effects for high velocities.)

Explanation / Answer

The 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)

Now, we just need to plug in the values of the various parameters.

E = 4.5*10^5 N/C
z = 1.602*10^-19 C
V = 18*10^3 V
m = 1.673*10^-27 kg

B = (4.5*10^5 N/C)/(1.856*10^6)

B = 0.2424 (N/m*A) = 0.2424 T

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