Question

A very large plane carries a surface charge density of +350 nC/m^2. 5 cm above and parallel to this plane is a thin disk with a diameter of 3 cm and a total charge of -2.5 nC. 5 cm above the disk, and on the axis of th disk, is a 1.5 nC point charge. Find the net force (magnitude and direction) acting on a proton halfway between the center of the disk and the 1.5 nC point charge.

Answer 1

Field due to large plane at proton is E1 = sigma/(2*eo) = 350*10^-9/(2*8.85*10^-12) = 19774 N/C

Field due to disc at the proton is E2 = (sigma/(2*eo))[1-(x/sqrt(x^2+R^2))]

sigma = -Q/A = 2.5*10^-9/(3.142*0.015^2) = -3.54*10^-6 C/m^2

x = 0.05/2 = 0.025 m

R = 1.5 cm = 0.015 m

then E2=((-3.54*10^-6)/(2*8.85*10^-12))*[1-(0.025/sqrt(0.025^2+0.015^2))]

E2 = -2.85*10^4 N/C

Field due to point charge at the proton is E3 = k*q/r^2 = (9*10^9*1.5*10^-9)/(0.025^2) = 2.16*10^4 N/C

then net field at the proton is E = E1+E2+E3

E3 is negative depending on the direction

E = 19774 -(2.85*10^4)-(2.16*10^4)

E = -3.0326*10^4 N/C

Force on proton is F = q*E = 1.6*10^-19*3.0326*10^4 = 4.85*10^-15 N towards the disk