1. What is the flux through the cylinder shown at left, inside the capacitor? a)E A, because the field strength is constant inside the capacitor b)0, because there is no charge inside the cylinder c)½ E A, because the field lines coming in on one side cancel some of those coming out on the other d)2 E A, because there are two cylinder ends between the plates there is twice the flux through the cylinder’s surface. - +

2. Correct Answer – B Gauss’ Law tells us that the total flux through a closed surface is equal to the charge inside the surface (divided by the electrical permitivity). Since the closed surface depicted is entirely within the insulating region between the plates, there is no charge inside it. Therefore the flux through it must be zero. How can we understand this in terms of flux lines? There clearly are flux lines entering through the left hand end of the cylinder. But everyone of those flux lines exits through the right hand end, and each line coming out cancels each line coming in.

3. What is the total flux passing through the cylinder shown?  is the surface charge density on each plate a)2  A, because that is the amount of charge inside the cylinder b)  A, because only one of the plates contributes to the flux through the cylinder c)½  A, because the negative charges on one plate cancel some of the charges on the other plate d)0, because there are no flux lines passing through the cylinder +  - 

4. Correct Answer – D If you look carefully at the diagram, you see there can be no flux lines passing through the closed surface of the cylinder. The curved surface of the cylinder is parallel to the flux lines, so none of the lines pass through this surface. In this cylinder the end surfaces are outside the capacitor, and we know that the flux lines, which cannot pass through the conducting plates, are not present outside the capacitor. So if the total flux through the closed surface is zero, Gauss’ Law says the total charge inside it must be zero. But we do know there is charge on the capacitor plates. The key here is that equally sized parts of both plates are inside the surface, so there must be an equal charge density on the two plates. Since one plate is positive and one plate is negative, the equal charge amounts cancel each other out.

5. + - Suppose we shrink the length of the negative plate to a half that of the positive plate. Where is the field strength greatest inside the capacitor? a)Close to the positive plate (longer plate) b)In the middle c)Close to the negative plate (shorter right) d)It is still the same everywhere inside the capacitor

6. Correct Answer – C A careful look at the diagram shows clearly that the flux lines are closer together closer to the smaller negative plate. Closer together flux lines correspond to stronger fields. The fact that the flux lines from the positive plate all end on the negative plate tells us that the two plates still have equal total charges Q. But since the area of the negative plate is smaller, the charge density on that plate is larger. Since electric field inside the capacitor depends directly on the charge density on the plates (see last pages of chapter 19) it follows that in the case of unequally sized plates the plate with greater charge density should have a greater electric field strength close to it.

7. + - In the case of the unequal plates, if we again draw a closed cylinder whose ends reach beyond the plates, what can we say about the total charge inside the cylinder? a)It is zero b)It is positive c)It is negative d)It is infinite

8. Correct Answer – C In the previous question we stated that the charge density on the smaller negative plate is greater than that on the larger positive plate. Since the cylinder has equal areas of the two plates within it, it follows there must be more negative charge inside the cylinder than there is positive charge. Therefore the total charge inside the cylinder is negative. What about the flux lines? There is still no flux through the ends of the cylinder, but the flux lines are no longer parallel to the curved surface of the cylinder. Since the flux lines all come diagonally into the cylinder from the outside it follows ther eis a net negative flux into the cylinder.

9. + - What is the field strength in the middle of the capacitor, equidistant from the two plates, for the flux lines drawn? Let E+ be the field strength close to the positive plate (the longer plate), and E- be the field strength close to the negative plate (the shorter plate). a)0 b)E- c)½ (E- + E+) d)E+ e)E- + E+

10. Correct Answer C – The flux lines are all drawn as straight in the illustration (this is not quite realistic, but is good enough for our purposes). Since they are straight, the change in field strength must be linear, so it follows that the field strength in the center must be the average of the field strength at the plates. What might be more realistic? We know field lines repel each other. When they are close together they repel more strongly, so the field lines can draw closer together more quickly near the positive plate where the field is weaker. This suggests the point where the field strength is halfway between the values at the plates might be a little closer to the positive (larger) plate.

11. +q F If a charge q is placed inside the capacitor, what is the amount of the force, F, propelling it towards the negative plate? (A is the area of the plate, r is the distance between the charge and the positive plate,  is the surface charge density on each plate,  is the electrical permittivity of free space). a)F = q E = q  /  b)F = k q Q /r 2 = k q  A/r 2 c)F = k q Q /r 2 + k q Q /r 2 = 2 k q  A/r 2 d)F = k q Q /r 2 – k q Q /r 2 = 0 +  - 

12. Correct Answer – A We know that a charge q placed in an electric field E experiences a force F = q E. Inside the plates of the capacitor with charge density , the field strength is E =  / 

13. Suppose we take a rectangular loop of wire and place it inside the capacitor as shown. Given that the field strength is constant everywhere in this capacitor, is there a way to move the wire so that the amount of flux passing through it changes? a)No, because the field is constant inside the capacitor. b)Yes, we just have to rotate the loop

14. Correct Answer – B It’s true that if we just move the loop up or down and back or forth that the flux through it won’t change. For each line that passes inside the loop, another line leaves, because the flux is constant inside the capacitor. But if we rotate the loop, matters are different. When the loop is parallel to the flux lines, they don’t actually pass through it, and the flux through it is zero. When it is perpendicular to the flux lines, they penetrate it maximally, and the flux through it is equal to E A where A is the area of the loop. So by rotating a loop we can change the flux through it, even when the field is constant. This will be important later on when we look at magnetic fields.