# Electrons in electric field

1. Jun 27, 2012

### nks27

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

work done by electric field = K.E of electrons

k.E = qV

3. The attempt at a solution

i cant figure how to work this out
Should potential energy also be considered?
Is my attempt at the question wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 27, 2012

### Infinitum

Hi nks27! Welcome to PF

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

3. Jun 28, 2012

### 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 dont get the answer.
Cud u plz correct my careless mistake if there's any?

4. Jun 28, 2012

### Saitama

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

5. Jun 28, 2012

### Infinitum

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,

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

6. Jun 28, 2012

### Infinitum

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

7. Jul 4, 2012

### nks27

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 doesnt give the right answer :(

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