Define the Capacitor With Dielectric

							











In this section, we consider the effect of filling the interior of a capacitor with a dielectric material. This effect was first investigated in 1837 by Michael Faraday. Faraday constructed two identical capacitors, filling one with a dielectric material and leaving the other with air between its plates. When both capacitors were connected to batteries with the same potential difference, Faraday found that the charge on the capacitor filled with the dielectric was greater than the charge on the capacitor with air between its plates. That is, the presence of the dielectric enables the capacitor to store charge. Since storage of charge for later discharge is one of the purposes for which we use capacitors, the presence of a dielectric can enhance the performance of a capacitor.
 

Figure 30-12a                    Figure 30-12b








 
 
The effect of filling a capacitor with dielectric depends on whether we do so with the battery connected (as in Faraday’s experiment) or disconnected. First we consider the situation as in Faraday’s experiment (Figure 30-12). A capacitor with capacitance C is connected to a battery of potential difference DV and allowed to become fully charged, such that the plates carry a charge q, as in Figure 30-12a. With the battery remaining connected, we then fill the interior of the capacitor with a material of dielectric constant ke, as in Figure 30-12b. The battery maintains the same potential difference DV across the plates.
 
If the potential differences in Figure 30-12a and 30-12b are the same, then the electric fields inside the capacitor must be the same. However, we would expect the presence of the dielectric to reduce the strength of the electric field. As Faraday concluded, the tendency of the additional charge that the battery delivers to the plates as the dielectric is inserted.
 
Let us assume that we are using a parallel-plate capacitor. With the capacitor empty, the electric field is given by Equation 30-3: E = s/Î0A. When the dielectric is present, the electric field is reduced by the factor 1/ke due to the presence o the dielectric, but the field is also changed because the plates now carry a charge q’, so the field is E’ = q’/keÎ0A. When the dielectric is present, the electric field is reduced by the factor 1/ke due to the presence of the dielectric, but the field is also changed because the plates now carry a charge q’, so the field is E’ = q’/keÎ0A. Since the fields must be equal, we can set E’ = E and conclude that q’ = keq.