The electric potential is the same everywhere inside a charged conductor. This is because the electric field is zero inside the conductor (see Ch. 15), and no work is done to move a charge through zero electric field. If no work is done, then the change in electric potential is zero, so it must be constant.
This result continues to hold at the surface of the conductor. At the surface of the conductor, the electric field is perpendicular to the surface. Therefore, if a charge moves along the surface, it moves perpendicular to the electric field, and perpendicular to the electric force. No work is done on an object that moves perpendicular to a force. So again we see that the electric potential must be constant inside and at the surface of a conductor.
Note that the electric field is zero inside a conductor, but the electric potential need not be zero, just constant.
A unit of energy common in atomic, nuclear, and particle physics is the electron volt, abbreviated eV. The electron volt is the energy released (or supplied) when an electron moves through a potential difference of 1V (-1V). Because one electron has a charge of -1.6×10-19C, and 1V = 1J/C, the eV is:
Imagine some fixed charges, and the electric field associated with them. Now imagine calculating the electric potential at lots of points around the charges. If we are clever, we can find locations around the charges where the electric potential is the same, and with a little thought, you should realize that these points form lines (or surfaces if we move to a 3 dimensional picture). These lines of equal electric potential are called equipotential lines, and they are another useful way to help us visualize the electric field. (Make sketches like those in Fig. 16.8) Notice that where the equipotential lines are always perpendicular to the electric field lines.
Capacitors are devices commonly used in electric and electronic circuits. In fact you may have seen something in the news last spring about a shortage of capacitors for cell phones -- a typical cell phone needs several dozen capacitors.
A capacitor has a rather simple construction: it looks like two identical conducting plates separated by a small gap, each plate can be connected to some device for generating a charge. We will consider only cases where the plates have equal amounts of oppositely signed charge. So, our standard picture of a capacitor is of two plates, each with area A, separated by a small gap, d, one with charge +Q and the other with charge -Q.
Capacitance, C, is defined as the charge on one conductor divided by the potential difference between them:
Note: We are running short on letters. The letter C stands for coulomb (charge) when used as a unit, and stands for capacitance when used as a symbol. You must determine which it is from the context. If C appears in an equation or other expression, it probably stands for capacitance. If C appears following a value, as in 12µC, it probably stands for coulomb.
A 3.0µF capacitor is connected to a 12V battery. What is the magnitude of the charge on each plate of the capacitor?
Use the definition of capacitance and solve for Q:
Q = CDV = (3.0µF)(12V) = 36µC
The parallel-plate capacitor was described above. A nice feature of this design (and the reason it is used as our standard capacitor) is that the capacitance can be determined by the construction. The capacitance depends on the area of the plates, A,the size of the gap, d, and the insulator between the plates. For now we will assume that the insulator is air, then:
From the equation for a parallel-plate capacitor we see that the capacitance will increase if the area of the plates is increased, or the gap between them is decreased. The capacitance will decrease if the area is decreased or the gap is increased. For example, if the spacing between the plates of a capacitor is reduced to half its original value, the capacitance will double. (This effect is used in computer keyboards and many types of sensors.)
Consider the Earth and a cloud layer 800m above the Earth to be the plates of a parallel-plate capacitor. (a) If the cloud layer has an area of 1.0km² = 1.0×106 m², what is the capacitance? (b) If an electric field strength greater than 3.0×106 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold?
(a) Assume the effective conducting are on the Earth matches the area of the cloud.
C = e0A/d = (8.85×10-12 C²/Nm²)(1.0×106 m²)/(800m) = 1.1×10-8 F.
A parallel plate capacitor has a uniform electric field between its plates, and zero field outside.
The potential difference when moving a distance d through a uniform E-field is DV = Ed.
In this case, the distance is gap between the plates of the capacitor, or the distance from the Earth to the clouds.
The charge needed to produce that potential difference is
Q = CDV = CEd = (1.1×10-8 F)(3.0×106 N/C)(800m) = 26 C.