Equipotentials These are lines through space which define regions of constant voltage or electric potential. The first rule to note with equipotentials, is that the surface of a conductor is always one equipotential. It has to be, because if there was a difference in potential along the surface, then charges would move through the conductor until the difference in potential was eliminated. Conductor Equipotenial Flux Line Conducting Surface

Shown is a parallel plate capacitor with the electric flux lines drawn in. If we were to draw the equipotential lines (lines along which the electric potential is constant) on this diagram, what lines should we draw? a)Horizontal lines, parallel to the flux lines b)Vertical lines, parallel to the plates c)Diagonal lines d)Concentric circles, centered on the midpoint between the plates +Q -Q

Correct Answer : B The lines are parallel to the plates. The plates are conductors, so the surfaces of the plates themselves are equipotentials, so it is not surprising that all of the equipotentials are parallel to the plates in a parallel-plate capacitor. This illustrates another rule, that equipotentials run perpendicular to field lines, which in this case are horizontal. Since the electric field is constant inside the capacitor, the distance between each pair equipotential is the same.

Energy stored in a Capacitor Besides being a device which stores charge, a capacitor also stores energy. We can say the electric field inside the capacitor contains energy and it would be interesting to know how much energy this amounts to. It will be useful to remember that a capacitor has the following properties Q = total charge on each plate (+ on one plate, - on the other) V = voltage difference between the two plates = E d E is the electric field strength in the region between the plates d is the distance between the plates A is the area of the plates C = capacitance of the plates = Q/V =  A/d =  o  A/d

The unit of Capacitance is the Farad named after our old friend Michael Faraday 1 Farad is the capacitance of a capacitor which will put 1 Coulomb of charge on its plate for each 1 Volt of potential difference between the plates. Most capacitors are measured in microFarads or picoFarads. A 1 F capacitor is unusual, expensive and can be dangerous! You’ll notice that checking one’s results by units is not as easy in electromagnetism as it was in mechanics. One always has constants like k and  with funny units which are hard to remember. Nevertheless if you are working the problem and have a book this is still an excellent way to check your answer.

Suppose we allow one of the positive charges on the left plate to fly across to the right hand plate where the negative charges are. What will happen? a)The charge on the capacitor and the energy stored in it will both be reduced b)The charge and the energy in the capacitor will both increase c)The charge will decrease on one plate and increase on the other, and the energy will be unchanged d)The charge will decrease but the energy stored will increase +q +Q -Q

Correct Answer – A The charge on both plates is reduced, because the positive plate lost one of its charges, and one of the negative charges on the other plate has had its field effectively “canceled out” by the arrival of a new positive charge. Since the charge on both plates is reduced, it follows that the strength of the electric field between them is also reduced, and it seems logical that a weaker electric field has less energy stored in it. We can check this by looking at the work done. The force on the charge generated by the field was pointing in the direction it moved, so the work done by the field on the charge was positive, and this means that the charge gained energy (in the form of kinetic energy) and the field lost energy. When the charge hits the plate on the other side the kinetic energy is dissipated as heat. So the field loses energy permanently.

+q What amount of energy is lost by the field when the charge flies across from one plate to the other? In other words, what work is done on the charge by the field in moving it across? a)W = E Q A b)W = Q V d c)W = Q q d d)W = E q d Here W = work done, E = electric field strength, Q = charge on plates, q = charge of the charge being moved across, A = area of plates, d = distance between plates +Q-Q

Correct Answer – D Remember that the work done by a force F moving an object a distance d is W = F d. But we also know that the force exerted by an electric field of strength E on a charge q is given by F = E q. Therefore the work done by the field on the charge is W = F d = E q d Note this q is the charge being moved across, not the total charge on the plates, Q.

+q How does the work done on the charge q moving across from one plate to the other, depend on the voltage difference between the plates. If W = E q d then which of the following is true? a)W = V Q b)W = V q d c)W = V q d)W = V

Correct Answer – C Recall that the relationship between electric field and potential difference or voltage is simple E = V/d The electric field is the voltage per unit distance, so V = E d Therefore W = E q d = V q

+q The best way to figure out the total energy stored in the field in the capacitor would be to calculate how much work would need to be done to move every positive charge on the left plate across to the right plate. After that the total charge on the plates would be zero, and the electric field and voltage would both be zero. All the energy would be gone. So after we had moved half of the charges across, what could we say about the work needed to move the next charge across compared to the work needed to move the first charge across? a)It would be just the same b)It would be half of the work on the first charge c)It would be twice the work on the first charge +Q -Q

Correct answer – B Once you have moved half the charges across from one plate to another, the total charge on the plates is now half what it was to begin with, which means the voltage between the plates is also halved, since V = Q/C (remember that C is constant). Since the electric field is only half what it was before, this means the force on the charge is half what is was before and therefore the work done is also halved, since W = F d = E q d If d and q are unchanged and E is halved, then so is W.

V Q U = area under line So if we think of draining the energy away slowly by moving the charge across, the change in energy  U when you decrease the charge on the plates by an amount  q is  U = W = V  q. This means the total energy on the plates, U, is the sum of of all these losses of energy, which means it is the sum of V  q, which is to say, the area under the curve on the plot of V versus Q. What is this result? a)U = ½ V Q b)U = V Q c)U = 2 V Q d)U = V Q 2

Correct Answer – A The area under the curve is easy in this case, since the curve is a straight line from Voltage = V, charge = Q (the initial state of the capacitor) to Voltage = 0, charge = 0 (the final state). The area under the line is half the area of a rectangle with one side of V Q U = area under line (V,Q) (0,0) length V and the other side of length Q. The area of the whole square = Q V, so the area of half the square is U = ½ Q V. Other ways of writing this are U = ½ C V 2 and U = ½ Q 2 /C, Since C = Q/V. If you can produce a given voltage with a battery, than the higher the capacitance the more energy your capacitor can store.

What does this tell us about the amount of energy which can be stored in an electric field? If U = ½ C V 2, then since V = E d we have U = ½ C E 2 d 2 Now we also know that C =  A /d so U = ½  A E 2 d Now if A is the area of the plates and d is the distance between them, than A d = V the volume of the space between the plates, which is to say, the volume of the space that our electric field occupies. So U/V = energy density in the electric field = ½  E 2 This result applies for any electric field, not just inside a capacitor