Question

In a cathode ray TV, the velocity selector controls the displacement of the ray in the vertical direction ∆y in order to project an image. The two conducting plates at LV creates the electric field in the y direction.

(a) 5 cm² parallel plates separated by 1 cm are at 3 V . Find Capacitance of the plates, σ of the plates, and E in the middle.

(b) Electric field in the −y direction is created by the plates. In the absense of any other fields, find the ∆y if the screen is 10 cm away. (Hint: Find the time ∆t, vy after ∆t, then ∆y.)

(c) Magnetic field in to the page is created by two coils. For E in (b), find B to achieve ∆y = 0.

Answer 1

Let us find out the capacitance,

(a) The capacitance is given by

$C= \epsilon _{0}\frac{A}{d}$

$C= \epsilon _{0}\frac{A}{d}= 8.85\times 10^{-12} \frac{F}{m}\frac{5\times 10^{-4}m^{2}}{1\times 10^{-2}m}= 0.44 \times 10^{-12}\, F$

now the charge´s can be determine by

$C= \frac{q}{V}$

$q= CV= 3V\times 0.44\times 10^{-12}F= 1.32\times 10^{-12} C$

hence the density is given by

$\sigma = \frac{ q }{A}= \frac{1.32 \times 10^{-12} C }{ 5 \times 10^{-4} m^{2} } = 0.26 \times 10^{-8} \frac{ c }{ m^{2} }$

The electric field is given by

$E= \frac{\sigma}{ \epsilon _{0}} = \frac{ 0.26\times 10^{-8}}{ 8.85\times 10^{-12}}= 0.03\times 10^{4}\frac{N}{c}$

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If we asume that the electron´s initial velocity is $v_{0}$, the time is given by

$t= \frac{v_{0}}{\Delta x}$

The electron´s vertical velocity is given by

$v_{y}= a_{y}t= \frac{qE}{m}t$

and $\Delta y$ is given by

$\Delta y= \frac{1}{2}a_{y}t^{2}= \frac{qE}{m}t^{2}$

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(c) In this case, the electric force balances the magnetic force, therefore

$qE= qVB$

$B= \frac{E}{V}$

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