Week 1 Solutions

Discussion Questions

2. The E field is 0 inside of the shell. Use Gauss's Law or by analogy to gravity or analogy with a spherical conductor, ... .

4. The beam will ionize the air and creates a conducting path to the film so that it can neutralize.

12. In winter there is less moisture in the air and so charge can build up.

14. The belt deposits charges onto the inside of the conducting shell. The charges then flow to the outer surface of the shell trying to get as far apart as possible. The amount of charge which can be stored on the dome is limited because when the charge becomes large, it can actually cause a breakdown of the surrounding air which allows the charge to flow off of the dome.

If we tried to add charge from the outside, we would run into trouble because the charges would have to overcome the intense electrical repulsion due to the charges already deposited onto the sphere. If we add charge from the inside this is not a problem because the charges which reside on the shell do not produce an electric field inside of the dome!

Problems

1. 1 electron has a charge of - 1.60 x 10**(-19) C. So, the number of electrons in - 1 micro C is

Number = [-10**(-6) C] / [ -1.6 x 10**(-19) C] = 6.2 x 10**12 electrons

The mass of an electron is 9.1x10**(-31) kg and so the added mass is

Mass = 6.2x10**12 x 9.1x10**(-31) = 5.6x10**(-18) kg

5. A proton has a charge of 1.6 x 10**(-19) C. The force between two protons separated by a distance of 10**(-14) m is repulsive and has a magnitude of

Force = 9.0 x 10**9 x [1.6 x 10**(-19)]**2 / [10**(-14)]**2 N


===> Force = 2.3 N

9. The force is 0, by symmetry

16. (i) Because the Coulomb force is an inverse square law and the charges have a ratio of 4 ===> the relative distance between the added charge and the other two charges must be sqrt(4):1 or 2:1.

(ii) Because the two charges have different signs, the third charge must be placed outside of the other two. It cannot lie in-between the other two.

(iii) The added charge must be placed at 1 meter from the negative charge and 2 meters from the positive charge in order for the forces to cancel.

25. The Electric field is given by E = F/q ===> E = 4x10**(-6) N / 5x10**(-9) C = 8x10**2 N/C

28. E = F/q = constant ==> F = q x constant. Since F = m x a ==> a = F/m = qE/m = constant ===> v = v(t=0) + (qE/m) t. Since we assumed that the particle was initially at rest, v(t=0) = 0 and v = (qE/m) t.

29.

37.

The field is perpendicular to the face of the capacitor plates (by symmetry). So,
• the flux out of the sides of the cylinder is 0.
• the flux out of the top (in the gap) and bottom (in the plate) of the cylinder is

  E x A + 0 x A = charge density x A / epsilon ===> E = charge density / epsilon

  Here A is the area of the end of the cylinder and we set the electric field in the plate to 0.

42.



i. Calculate the components of the field:

• Ex=9.0x10**9 x [10**(-6) C]/(0.2 m)**2 - 9.0x10**9 x 10**(-6) C / (0.2 m)**2 = 0

• Ey=-9.0x10**9 x [3x10**(-6) C/(0.3 m)**2 = 3x10**5 N/C

43. By symmetry, we expect that the electric field due to the cylinder will be in the direction perpendicular to the axis of the cylinder (see example 17.8). Since the charge is positive, we expect the field lines to point away from the cylinder. Construct the Gaussian surface as follows. As in example 17.8, make a cylinder of length L and radius r > radius R of the conducting cylinder.

• Because the field is radial, there is no flux out the ends of the cylinder.
• Because the field is radial, we have
  E x (2 x pi x r x L) = (charge density x 2 x pi x R x L) / epsilon
  The field is then E = charge density x (R/r) / epsilon in the outward radial direction.