Finding the Minimum Kinetic Energy of Electrons in an Electric Field

[nks27]

Homework Statement

Two metal plates, PQ AND RS are separated by a distance of 15 mm.
PQ maintained at potential of +100V relative to RS.
Beam of electrons of different kinetic energies directed a slit ,on plate PQ, at angle of 60 degrees to plate.
To find the K.E of electrons that 'just' reach the plate RS.

Homework Equations

work done by electric field = K.E of electrons

k.E = qV

The Attempt at a Solution

i can't figure how to work this out
Should potential energy also be considered?
Is my attempt at the question wrong?

 

[Infinitum]

Hi nks27! Welcome to PF

Homework Statement

Two metal plates, PQ AND RS are separated by a distance of 15 mm.
PQ maintained at potential of +100V relative to RS.
Beam of electrons of different kinetic energies directed a slit ,on plate PQ, at angle of 60 degrees to plate.
To find the K.E of electrons that 'just' reach the plate RS.

Homework Equations

work done by electric field = K.E of electrons

k.E = qV

The Attempt at a Solution

i can't figure how to work this out
Should potential energy also be considered?
Is my attempt at the question wrong?

I believe you want to use the energy conservation principle. So yes, you do need to include potential energy...

 

[nks27]

i the use this formula

work done
against electric = K.E + P.E
field
K.E = work done - P.E

= qV - q/(4πεₒr)

but i don't get the answer.
Cud u please correct my careless mistake if there's any?

 

[Saitama]

i the use this formula

work done
against electric = K.E + P.E
field

From where did you get potential energy?
Instead using the energy conservation, you can use the equations of motion too.

 

[Infinitum]

i the use this formula

work done
against electric = K.E + P.E
field
K.E = work done - P.E

= qV - q/(4πεₒr)

but i don't get the answer.
Cud u please correct my careless mistake if there's any?

Umm no. How did you get q/(4πεₒr)??


The kinetic energy change will result in potential energy. Now, for the minimum kinetic energy you need the final velocity to have no x component(assuming vertical plates). So you have,

$$\frac{1}{2}m(v_x)^2 +\frac{1}{2}m(v_y)^2 = qV + \frac{1}{2}m(v_y)^2$$

 

[Infinitum]

From where did you get potential energy?
Instead using the energy conservation, you can use the equations of motion too.

I would prefer conservation of energy, but this method works too

 

[nks27]

Umm no. How did you get q/(4πεₒr)??


The kinetic energy change will result in potential energy. Now, for the minimum kinetic energy you need the final velocity to have no x component(assuming vertical plates). So you have,

$$\frac{1}{2}m(v_x)^2 +\frac{1}{2}m(v_y)^2 = qV + \frac{1}{2}m(v_y)^2$$

sorry for late post.
substituting v_x = vcos60
and v_y = vsin60

q= 1.6 X 10^-19 and V= 100 V


i get v^2 to be 1.41 X 10^14

using it to calculate the k.E doesn't give the right answer :(


the " q/(4πεₒr) " was for electric potential .