## SI123 Group Title The electric field between the 2 parallel plates of an oscilloscope is 1.2 x 10^5 V/m. If an electron of energy 2 keV enters at right angles to the field, what will be its deflection if the plates are 1.5 cm long? one year ago one year ago

1. Mashy Group Title

The deflection angle is required?

2. SI123 Group Title

deflection angle is not mentioned in the book

3. Mashy Group Title

then what is asked? what you mean find deflection??

4. SI123 Group Title

|dw:1361792988295:dw| Y is the deflection that has to be calculated

5. DLBlast Group Title

the angle should be 37

6. DLBlast Group Title

|dw:1361855358005:dw|for angles 180=x+y+z

7. DLBlast Group Title

oh 45 is the other number it could be

in this case, since it is constant, uniform acceleration due to the electric field, by Newton's second law, $$\Sigma \vec F= q \vec E = m \vec a$$ where q and m of the electron are known constants. so, $$\vec a=\frac{q \vec E}{m}$$, which will be different in direction of the electric field, as expected, now, the original speed of the electron can be calculated from: $$\frac{1}{2}mu^2= 2KeV$$, so $$u=\sqrt{ \frac{4KeV}{m}}$$ by the equations of motion, $$\vec S_x=\vec u_x t + \frac{1}{2} \vec a_x t^2$$, but $$\vec u_x=\vec u$$ and $$\vec a_x =0$$ and $$\vec S_x =1.5cm$$ so, $$t=\frac{1.5}{\sqrt{ \frac{4KeV}{m}}}$$ now, for $$\vec S_y=\vec u_y t + \frac{1}{2} \vec a_y t^2$$, but this time,$$\vec u_y=0$$, $$\vec a_y = \vec a = \frac{q \vec E}{m}$$ so, $$\vec S_y= \frac{1}{2} \vec a_y t^2$$ sub it all in to get the displacement along the y axis. Do you need further help?