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.