Q 25 Hsc 2002 (1 Viewer)

[206wrc]

Question 25 (6 marks)
A pair of parallel metal plates, placed in a vacuum, are separated by a distance
of 5.00 × 10−3 m and have a potential difference of 1000 V applied to them.

(a) Calculate the magnitude of the electric field strength between the plates.

(b) Calculate the magnitude of the electrostatic force acting on an electron between
the plates.

(c) A beam of electrons is fired with a velocity of 3.00 × 106 m s−1 between the
plates as shown. A magnetic field is applied between the plates, sufficient to
cancel the force on the electron beam due to the electric field.
Calculate the magnitude and direction of the magnetic field required between
the plates to stop the deflection of the electron beam.


I'm having a problem with part b) and c)

Part b)
F=Eq is the equation but what is 'q'? I know its charge but how do you get it?

Part c)
I need a whole explanation

Thanks

 

[JhK89]

im pretty crap at physics but ill give it a go..

b) q is the charge of an electron its given to u on the data sheet in the HSC

its 1.6x10^-19

c) the Magnintude of the field is calculated by deriving the eqns F=qvBsin@ and F=Eq from b).

Eq=qvBsin@
E=VbSin@
sin@=1 because it perepndicular
E=vB
5=3x10^6 x B

B= 5/3x10^6

B= 1.6x10^-6
( i think )

and using the right hand palm rule, the direction is into the page by the flow of electrons hope this helps if im right..

 

[Mattius]

Im fairly sure your wrong, but not far off.

Stateing that force = electron charge i think is incorrect.

would it not be F=eq

hence F= 1.602x10^-19 x 20000 (part a)

=-3.204^-14
as its a force i just say its positive

C) As forces are equal they can be equated
F(electric) = F(magenetic)
Therefore

-3.204^-14 = qvB

-3.204^-14/qv = B

B = 0.0666 tesla
After looking up the diagram the force needs to opose the electric field. Hence the magnetic field would be into the page.

Final answer = 0.06666 Tesla into the page.

 

[JhK89]

i am very sorry

 

[angmor]

qvB = mv2 / r

just remember that rule,. especially when asking about Thomson's charge to mass ratio