Electrical Energy and Voltage

Question

Two parallel plates are separated by a distance of 12.3 mm are connected to a 9 volt battery. A proton is emitted from positive plate with an initial speed of 1.62 × 10^4 m/s.

I need A and E answered.

(A) What is the speed of the proton when it is at a potential of 1.91 volts?

(B) At what distance from the negative plate is the proton when it is at a potential of 1.91 volts?

2.61 mm

(C) What is the magitude of the electric field between the plates?

731.71 V/m

(D) What is the electric potential of the proton when it is 7.03 mm from the negative plate?

5.14 V

(E) How fast is the proton moving when it is 7.03 mm from the negative plate?

Answer 1

(A)   When the proton is at 1.91 volts, it experienced a potential drop of 9 – 1.91 = 7.09 volts.

 

This loss of potential energy results in an increase in kinetic energy which is given by

ΔKe = ΔV x Q = 7.09 volts x charge of proton.

Final Ke = initial Ke + loss of PE = ½ x m x vi^2 + ΔV x Q joules = ½ x m x vf^2

Solving for vf^2 gives vi^2 + 2 x ΔV x Q / m

Vf = √ (vi^2 + 2 x ΔV x Q / m)

I got a numerical result of 4.03 x 10^4 m/s

I think you can work out (E) yourself. (Hint: calculate the potential at 7.03 mm from negative plate and use the approach above.