trwo questions i cannot figure out 1) Two 3.0 cm -diameter disks face each other

Question

trwo questions i cannot figure out

1) Two 3.0 cm -diameter disks face each other, 1.9 mm apart. They are charged to ±20nC. What is the electric field strength between the disks? 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

2)What are the strength and direction of the electric field at the position indicated by the dot in the .figure(Figure 1) ? Give your answer in component form. Assume thatx-axis is horisontal and points to the right, and y-axis points upward.figure(Figure 1) ?Ex?,Ey? What is the magnitude of the electric field at the position indicated by the dot in the figure.

Explanation / Answer

1)

Electric field between disks is E = sigma/eo = (Q)/(A*eo)

A = pi*r^2 = 3.142*(0.03/2)^2 = 7.06*10^-4 m^2

E = (20*10^-9)/(7.06*10^-4*8.85*10^-12) = 3.2*10^6 N/C

using q*E*S = 0.5*m*v^2

1.6*10^-19*3.2*10^6*1.9*10^-3 = 0.5*1.67*10^-27*v^2

v = 1.07*10^6 m/s

2)field due to 10 nC is E1y = k*q1/r1^2 = (9*10^9*10*10^-9)/(0.03^2) = 10*10^4 N/C

Field due to -5 nC is E1x = -k*q2/r2^2 = (-9*10^9*5*10^-9)/0.05^2 = -18*10^3 N/C

Field due to -10 nC is E2 = k*q3/(0.05^2+0.03^2) = (9*10^9*10*10^-9)/(0.05^2+0.03^2) = 2.64*10^4 N/C

E2x = E2*cos(theta) = 2.64*10^4*(0.05)/sqrt(0.05^2+0.03^2) = 2.26*10^4 N/C

E2y = E2*sin(theta) = 2.64*10^4*0.03/sqrt(0.05^2+0.03^2) = 1.35*10^4 N/C

x-component field is Ex = (-1.8*10^4)+(2.26*10^4) = 0.46*10^3 N/C

Y-component field is Ey = (-10*10^4)+(1.35*10^4) = -8.65*10^4 N/C

magnitude is E = sqrt(Ex^2+Ey^2) = sqrt(0.46^2+8.65^2)*10^4 = 8.66*10^4 N/C

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