Fall 97 Test 1, P. 2

2102 Fall 97 Test 1

Fall 97 Test 1, P. 5 5. (10 points) A spherical rubber balloon has a charge uniformly distributed over is surface. The balloon is slowly inflated and remains spherical during inflation. a) At a point well away from the balloon, does the electric field increase, decrease, or stay the same as the balloon inflates. Explain why. Use only words in complete sentences to explain your answer; no formulas, symbols or diagrams. At a point well away from a spherical symmetric charge, the electric field is independent of the size of the sphere and therefore will stay the same as the balloon inflates. A point charge, a spherical shell of charge, a charge distributed uniformly throughout a spherical volume all produce the same field if the total charge is the same. The easiest way to see this is with Gauss law. A spherical gaussian surface through a point away from the surface will have the same flux going through it no matter what the radius of the balloon on the inside. b) At the outer surface of the balloon, does the electric field increase, decrease, or stay the same as the balloon inflates. Explain why. Use only words in complete sentences to explain your answer; no formulas, symbols or diagrams. Since the distance from the center to the outer surface of the balloon increases as the balloon inflates, the electric field will decrease as the balloon inflates. Again using Gauss law with a spherical gaussian surface incrementally outside the surface, the flux which is field times area must stay the same since the charge inside is constant. Since the area increases, the electric field must decrease in order for flux to be constant. c) At a point inside the balloon, does the electric field increase, decrease, or stay the same as the balloon inflates. Explain why. Use only words in complete sentences to explain your answer; no formulas, symbols or diagrams. By Gauss law, the field everywhere inside a spherical shell of charge is zero. Take a spherical gaussian surface with a radius less than that of the balloon. There is no charge inside, therefore the flux is zero meaning there can be no electric field. The field inside stays the same (equal t o zero) as the balloon inflates.

Fall 97 Test 1, P. 6 6. (10 points) Describe how you would use Gauss law to calculate the electric field at point P between the two charged infinitely long cylinders shown with linear charge densities of λ 1 and λ 2. Describe: the gaussian surface you would use (size, shape, orientation). how you would calculate the flux through your gaussian surface. how you would determine the charge inside your gaussian surface. Use only words and proper English in your answer; no formulas, symbols or diagrams. In this case the gaussian surface will be a cylinder through the point, concentric with the charged cylinders. It will be an arbitrary length. Since the charged cylinders are infinitely long, the electric field between the cylinders is radial and perpendicular to the axis. This means that the electric field is perpendicular to the curved surface of the cylinder. P λ 1 λ 2 The flux through this part of the gaussian surface is simply the electric field times the area of the surface or the circumference times the length. The flux through either end cap of the gaussian cylinder is zero since the electric field is parallel to the surface. On the other side of Gauss law we have to calculate the charge inside the cylinder. This is simply the linear charge density on the inner cylinder times the arbitrary length. Equating these two, the arbitrary length cancels out leaving electric field equal to the linear charge density divided by the permittivity of free space times the circumference of the gaussian cylinder.

Fall 97 Test 1, P. 7 7. You are given a parallel plate capacitor with square plates of area A, and separation d, in a vacuum. Each plate has a charge of magnitude q. There is no battery nor any other source of charge connected to the plates. If the plate area is doubled, explain what happens to: the surface charge density, the electric field between the plates, the capacitance. Explain the physics behind your answers. Use only words in complete sentences. Do not use formulas, symbols, or diagrams in your answer. The surface charge density, being the charge divided by the area, simply decreases by a factor of 2 if the area is doubled. The charge that is present must now spread itself out over an area that is twice as large. The electric field between the plates, therefore, will decrease by a factor of two since it is directly proportional to the surface charge density. The capacitance will now be a factor of two larger since the charge on the plates is the same but the potential is a factor of two lower due to the lower electric field. Since the capacitance is the ratio of the charge to the potential applied to the plates, and the potential is a factor of two lower, the capacitance must be a factor of two larger.