Question

An electron initially has a speed 19.4 km/s along the x-direction and enters an electric field of strength 24 mV/m that points in the y-direction and has a width of 18.5 cm. How far is the deflection?

Answer 1

$$d=vt$$

$$t=\frac{d}{v}=\frac{0.185}{19400}=9.536\cdot 10^{-6}\ s$$

The electrostatic force:

$$F=qE$$

The acceleration:

$$a=\frac{F}{m}=\frac{qE}{m}$$

The deflection is

$$h=0.5at^2=\frac{qE}{2m}t^2$$

$$h=\frac{(1.6\cdot 10^{-19})(0.024)}{2(9.11\cdot 10^{-31})}(9.536\cdot 10^{-6})^2=0.192\ m$$