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as shown in figure, potential difference V exists between two parallel plates (a,b)...

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kibau | eNoter

Posted May 30, 2013 at 4:15 PM via web

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as shown in figure, potential difference V exists between two parallel plates (a,b) that have a small hole. Uniform electric field `vecE`  and magnetic flux density `vecB`  exist in the region above the plates, and are perpendicular to each other. The direction of `vecE` is parallel to this page and plate b. The direction of `vecB` is perpendicular to this page, from back to front. A positively charged particle of charge q and mass m, initially at rest in the hole of plate a, is accelerated by potential difference V so that it enter the region above the plate perpendicularly to `vecE` and `vecB` and travels straight through the region``

what is the ratio of the magnitude E of the electric field to the magnitude B of the magnetic flux density E/B?

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted September 13, 2013 at 1:21 PM (Answer #1)

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The particle is accelerated from rest by the potential difference U until it reaches a speed v when it exits the space between plates (a,b).

Kinetic energy when particle leaves the plates = Electric potential energy given between plates

`(m*v^2)/2 =q*U`

`v = sqrt((2*q*U)/m)`

Now, in the region where there exist both fields E and B the Lorentz force on the charge q is (v is perpendicular to both E and B):

`F = q*(E +vB)`

Condition of non-deflection is

`F =0`   or equivalent `E = -vB`    which means that in absolute

value the ratio E/B is

`E/B = v = sqrt((2*q*U)/m)`

Observation:

In practice this experiment is used to determine the rapport (q/m) for the electron or any given particle of charge q and mass m, by measuring the fields E and B under zero deflection conditions.

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