Page 376 - Physics Coursebook 2015 (A level)

Cambridge International A Level Physics
364
earth
Figure 23.8 Equipotential lines in a uniform electric field.
We can extend the idea of electric potential to measurements in electric fields. In Figure 23.9, the power supply provides a potential difference of 10 V. The value of the potential at various points is shown. You can see that the middle resistor has a potential difference across it of (8 − 2) V = 6 V.
+10 V
4Ω
+8 V
12Ω +2 V 4Ω
0V
+10 V 0V
field lines
++++
+Q V
Distance d
Figure 23.10 The potential changes according to an inverse
law near a charged sphere.
able to see how this relationship parallels the equivalent formula for gravitational potential in a radial field:
GM φ= − r
Note that we do not need the minus sign in the electric equation as it is included in the charge. A negative charge gives an attractive (negative) field whereas a positive charge gives a repulsive (positive) field.
We can show these same ideas by drawing field lines and equipotential lines. The equipotentials get closer together as we get closer to the charge (Figure 23.11).
+ Q
Figure 23.11 The electric field around a positive charge. The dashed equipotential lines are like the contour lines on a map; they are spaced at equal intervals of potential.
To arrive at the result above, we must again define our zero of potential. Again, we say that a charge has zero potential energy when it is at infinity (some place where it is beyond the influence of any other charges). If we move towards a positive charge, the potential is positive. If we move towards a negative charge, the potential is negative.
equipotential lines
Figure 23.9 Changes in potential (shown in red) around an electric circuit.
Energy in a radial field
Imagine again pushing a small positive test charge towards a large positive charge. At first, the repulsive force is weak, and you have only to do a small amount of work. As you get closer, however, the force increases (Coulomb’s law), and you have to work harder and harder.
The potential energy of the test charge increases as you push it. It increases more and more rapidly the closer you get to the repelling charge. This is shown by the graph in Figure 23.10. We can write an equation for the potential V at a distance r from a charge Q:
V=Q 4πε0r
(This comes from the calculus process of integration, applied to the Coulomb’s law equation.) You should be