Answer to Question #91753 in Electricity and Magnetism for shannon

Question

Two metal plates are squares of 12 cm by 12 cm, separated by 1.0 mm of air. Wires connected to each plate are also connected to a 9.0 V battery. After a long time, the battery is disconnected and the two plates are arranged to be perpendicular instead of parallel and almost touching each other along one edge (an “L” shape). A proton (charge +1.6 x 10-19 C, mass 1.67 x 10-27 kg) is released from rest from the middle of the positive plate. When and where does the proton hit the negative plate?

Answer 1

Initially the plates had a charge of

$$Q=CV=\frac{\epsilon_0b^2V}{d}.$$

The electric field of the both plates will be

$$E=E_-=E_+=\frac{Q}{2a^2\epsilon_0}=\frac{\epsilon_0V}{2d}.$$

One plate is positively charged and the other is negative, 90 degrees to each other. The total electric field according to the superposition principle thus will be

$$E_t=\sqrt{E_+^2+E^2_-}=\frac{\epsilon_0V}{\sqrt{2}d}.$$

The force acting on the proton will be

$$F=Eq=\frac{q\epsilon_0V}{\sqrt{2}d},$$

and according to Newton's second law, the acceleration of the proton will be

$$a=\frac{F}{m}=\frac{q\epsilon_0V}{\sqrt{2}md}.$$

The distance from the centre of negative to the centre of positive plate is

$$l=\frac{b}{\sqrt{2}},$$

hence the time requires to cover that distance with such acceleration is

$$t=\sqrt{\frac{2l}{a}}=\sqrt{\frac{2mbd}{q\epsilon_0V}}=0.177\text{ s}.$$

The proton will hit the negative plate at its centre.