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By the end of this section, you will be able to:
  • Describe the effects a dielectric in a capacitor has on capacitance and other properties
  • Calculate the capacitance of a capacitor containing a dielectric

As we discussed earlier, an insulating material placed between the plates of a capacitor is called a dielectric. Inserting a dielectric between the plates of a capacitor affects its capacitance. To see why, let’s consider an experiment described in [link] . Initially, a capacitor with capacitance C 0 when there is air between its plates is charged by a battery to voltage V 0 . When the capacitor is fully charged, the battery is disconnected. A charge Q 0 then resides on the plates, and the potential difference between the plates is measured to be V 0 . Now, suppose we insert a dielectric that totally fills the gap between the plates. If we monitor the voltage, we find that the voltmeter reading has dropped to a smaller value V . We write this new voltage value as a fraction of the original voltage V 0 , with a positive number κ , κ > 1 :

V = 1 κ V 0 .

The constant κ in this equation is called the dielectric constant    of the material between the plates, and its value is characteristic for the material. A detailed explanation for why the dielectric reduces the voltage is given in the next section. Different materials have different dielectric constants (a table of values for typical materials is provided in the next section). Once the battery becomes disconnected, there is no path for a charge to flow to the battery from the capacitor plates. Hence, the insertion of the dielectric has no effect on the charge on the plate, which remains at a value of Q 0 . Therefore, we find that the capacitance of the capacitor with a dielectric is

C = Q 0 V = Q 0 V 0 / κ = κ Q 0 V 0 = κ C 0 .

This equation tells us that the capacitance C 0 of an empty (vacuum) capacitor can be increased by a factor of κ when we insert a dielectric material to completely fill the space between its plates . Note that [link] can also be used for an empty capacitor by setting κ = 1 . In other words, we can say that the dielectric constant of the vacuum is 1, which is a reference value.

Figure a shows a capacitor connected in series with a switch and a battery. The switched is closed and there is a voltmeter across the capacitor, showing the reading V0. The plates of the capacitor have charge +Q0 and –Q0. Figure b shows the same circuit, with the switch open. This is labeled Step 1. The space between the plates of the capacitor is grey colored, indicating the presence of a dielectric. This is labeled Step 2. The positively charged plate has negative signs on the inside, labeled –Qi. The negatively charged plate has positive signs on the inside, labeled plus Qi. The voltmeter shows the reading V, which is less than V0.
(a) When fully charged, a vacuum capacitor has a voltage V 0 and charge Q 0 (the charges remain on plate’s inner surfaces; the schematic indicates the sign of charge on each plate). (b) In step 1, the battery is disconnected. Then, in step 2, a dielectric (that is electrically neutral) is inserted into the charged capacitor. When the voltage across the capacitor is now measured, it is found that the voltage value has decreased to V = V 0 / κ . The schematic indicates the sign of the induced charge that is now present on the surfaces of the dielectric material between the plates.

The principle expressed by [link] is widely used in the construction industry ( [link] ). Metal plates in an electronic stud finder act effectively as a capacitor. You place a stud finder with its flat side on the wall and move it continually in the horizontal direction. When the finder moves over a wooden stud, the capacitance of its plates changes, because wood has a different dielectric constant than a gypsum wall. This change triggers a signal in a circuit, and thus the stud is detected.

Practice Key Terms 1

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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