Question

A proton is released from rest at the positive plate of a parallel plate capacitor. The charge per unit area on each plate is σ=1.8e^-7 C/m² , and the plates are separated by a distance of 1.5e^-2 m.

a. What is the magnitude of the electric field between the two plates?

b. What is the potential difference between the two plates?

c. The line connecting A and C is perpendicular to the electric field lines. The distance between A and C is 0.7e^-2m. What is the potential difference between A and C, specifically V_A-V_c=?

d. Points A and B are along an electric field line, with a distance of 0.8^-2m between them. What is the potential difference between A and B, specifically V_B-V_A=?

e. What is the potential difference between B and C, specifically V_B-V_C=?

f. What is the speed of the proton when it reaches the negative plate? Solve this problem using conservation of energy.

g. Solve (f) using kinematics equations.

Answer 1

Given sigma = 1.8 times 10^-7 c/m^2 d = 1.5 times 10^-2 m (a) parallel plate capacitor From Gauss law E = phi/elementof_0A = sigma/elementof_0 = 1.8 times 10^-7/8.854 times 10^-12 N/C Tildeequal 0.203 times 10^5 N/C Tildeequal 2.03 times 10^4 N/C Ans (b) Potential difference between two plates = dE = 1.5 times 10^-2 m times 2.03 times 10^4 N/C = 3.045 times 10^2 Joule Tildeequal 3.05 times 10^2 Joule Ans (c) A C line is perpendicular to electric field lines. As parallel electric fields, so all points on perpendicular to electric field same potential. Therefore Potential A = Potential C therefore Potential A = Potential C therefore 0.7 times 10^-2 m apart A and C Potential difference = potential A - Potential C = 0 Ans (d) d_AB = 0.8 times 10^-2 m v_AB = d_AB E = 0.8 times 10^-2 times 2.03 times 10^4 Joule = 1.62 times 10^2 Joule Ans (e) v_BC = same as