Question

Two 3.5-cm-diameter disks face each other, 2.0 mm apart. They are charged to ± 11 nC . a) What is the electric field strength between the disks?
Express your answer in newtons per coulomb.

b) A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?
Express your answer in meters per second.

Answer 1

Answer:

a) 1.29*10^6 N/C

b) 0.703 *10^6 m/s

Explanation:

This is a parallel plates capacitor. In a parallel plates capacitor the electric field depends on the charge of the disks, its area and the vacuum permisivity (Assuming there is no dielectric) and can be found using the expression:

$E = \frac{Q}{A*e_0} =\frac{11*10^{-9}C}{(\frac{1}{4}\pi*(0.035m)^2)*8.85*10^{-12}C^2/Nm^2} = 1.29 *10^6 N/C$

For the second part, we use conservation of energy. The change in kinetic energy must be equal to the change in potential energy. The potential energy is given by:

$PE = V*q$

V is the electric potential or voltage, q is the charge of the proton. The electric potential is equal to:

$V = -E*d$

Where d is the distance to the positive disk. Then:

$\frac{1}{2}mv_1^2 +V_1q = \frac{1}{2}mv_2^2 +V_2q\\\frac{1}{2}m(v_1^2 - v_2^2)=(V_2-V_1)q = (r_1-r_2)Eq|r_2 = 0m, v_2=0m/s\\v_1 = \sqrt{2\frac{(0.002m)*1.29*10^6 N/C*1.6*10^{-19}C}{1.67*10^{-27}kg}}= 0.703 *10^6 m/s$