I need help with this homeworks ASAP PLEASE . Thank you.. 1) An infinite noncond
ID: 1436094 • Letter: I
Question
I need help with this homeworks ASAP PLEASE . Thank you..
1) An infinite nonconducting sheet has a surface charge density = 0.13 µC/m2 on one side. How far apart are equipotential surfaces whose potentials differ by 75 V?
2) In the figure particles with charges q1 = +3e and q2 = -15e are fixed in place with a separation of d = 25.0 cm. With V = 0 at infinity, what are the finite (a) positive and (b) negative values of x at which the net electric potential on the x axis is zero?
3)How much work is required to set up the four-charge configuration of the figure if q = 2.55 pC, a = 63.5 cm, and the particles are initially infinitely far apart and at rest?
4) In the figure a charged particle (either an electron or a proton) is moving rightward between two parallel charged plates separated by distance d = 5.60 mm. The plate potentials are V1 = –72.0 V and V2 = –49.0 V. The particle is slowing from an initial speed of 81.0 km/s at the left plate. (a) Is the particle an electron or a proton? (b) What is its speed just as it reaches plate 2?
Explanation / Answer
1) The electric field can be found from Gauss' law
E = s/e0. Where s = surface charge density
The field is perpendicular to the sheet so the E field is just the derivative with respect to the normal to the sheet, call that direction x. Hence
E = -dV/dx. So integrate to get V
V = - sx/e0
Now DV = 75 volts which is the difference between two equipotential surface so,
DV = E(x2) - E(x1) = s/e0*[x1 - x2] = s/ e0 Dx. Solve for Dx
Dx = e0*DV/s.
Dx = 8.85*10^-12*75/0.13*10^-6
= 5.10*10^-3m
2) V = kq1/r1 + kq2/r2
Factor out k since it appears in both terms
0 = 3e/x -15e/(0.25-x) where x is the point on the x axis between the two charges that are 0.3m apart.
3e/x = 15e/(0.25-x) => 3e*(0.25-x) = 15e*x => 0.75e -3ex = 15ex =>0.75e = 18ex => x = 0.04m
Now look for V = 0 on negative x axis. Let x be the distance to the left of the origin
0 = 3e/x -15e/(0.25+x) => 3e*(0.25+x) = 15e*x => 0.75e +3ex = 15ex =>0.75e = 12ex => x = 0.6m left from the origin.
V = 0 at x = 4cm and x = -60cm
3) k = 9*10 Nm²C²
Let the square ABCD have +q, -q, +q and -q at A, B, C, D respectively.
No energy required to bring q at A
Energy required to bring -q at B = -{(kq²)/a} -------------------- (1)
Energy required to bring q at C = -{(kq²)/a} +{(kq²)/(a*2)}-------------------- (2)
Energy required to bring -q at D = -2*{(kq²)/a} +{(kq²)/(a*2)}-------------------- (3)
Total energy = {(2) - 4}*{(kq²)/a} = -[{2.635*9*(2.55²)*(10¹)}/(0.635)] = -1.17*1013 J
4) Consider the work-energy theorem. You are correct that the particle is a proton, because it is slowing down, meaning that it is losing kinetic energy as it moves to greater potential.
The amount of work done on the charge from -70V to -50V is:
W = (-72V+49V)*q
W = (-23V)*q
q = 1.6e-19C
W = (-23V)*(1.6e-19C)
W = -3.68e-18J
This is the amount of kinetic energy that the proton will have lost when it gets to the other plate.
initial K = (1/2)mv^2
= (1/2)*(1.673e-27kg)*(81 km/s)^2
= 5.47e-24J
final K = 5.47e-24J - 3.68e-18J
= 3.67e-18J
final velocity:
v = sqrt( 2*K/m )
= 66.29km/s <===there's your answer