Capacitors with Dielectrics

A dielectric is a nonconducting material that increases the capacitance of a capacitor when inserted between its plates. When a dielectric slab is inserted in a uniform electric field, positive portions of the molecules are shifted in the direction of the electric field while the negative portions are shifted in the opposite direction. Hence, the applied electric field polarizes the dielectric, which in turn creates an induced positive surface charge density on the right face and a negative surface charge density of the same magnitude on the left face. These induced charges give rise to an induced electric field that opposes the external electric field. Therefore, the net electric field decreases. Thus, the capacitance increases.

If a dielectric fills the space between the two plates, the capacitance increases by a constant, k. We call this constant the dielectric constant. Thus, C=kCo, where C is the capacitance after the insertion of the dielectric and Co is the capacitance in the absence of the dielectric. Because the charge on a capacitor does not change with the insertion of a dielectric, the voltage drop across the capacitor must decrease. Thus, V=Vo / k.

For any given separation between the plates of the capacitor, the maximum voltage that may be applied to the capacitor without a resulting discharge depends on the dielectric strength of the dielectric between the plates. The dielectric strength for air is 3 MV/m. If the field strength of a medium exceeds the dielectric strength, then the insulating properties will break down and the medium will then begin to conduct. Most insulating materials have dielectric strengths greater than air's . Such dielectrics provide the following advantages:
1. A dielectric increases the capacitance of a capacitor.
2. A dielectric increases the maximum operating voltage of a capacitor.
3. A dielectric may provide mechanical support between the conducting plates.

According to the formula for the energy of a capacitor, one can also see that this energy decreases when a dielectric is inserted between the plates of a capacitor.

Energy (U)= CV2/2 = QV/2 = Q2/2C

Problems with Dielectrics

1.  A parallel-plate capacitor has a capacitance of 5 uF.  If a piece of Pyrex glass is placed between the

plates, what is the new capacitance?  (The dielectric constant of Pyrex glass 5.6.)


    C = k Co
    C = (5.6) (5 uF)
    C = 28 uF


2.  A parallel-plate capacitor has a capacitance of 10 uF and a potential difference of 15 V.  If a sheet

of paper is placed between the plates, what is the new potential difference across the capacitor?

(The dielectric constant of paper is 3.7.)


    V = Vo / k
    V = 10 V / 3.7
    V = 2.7 V


3.  A parallel-plate capacitor has a capacitance of 20 uF.  The capacitor is charged to a potential

difference of 10 V.  If a slab of Pyrex glass is placed between the plates of the capacitor:
    (a)  What is the new capacitance?
    (b)  What is the new potential difference of the capacitor?
    (c)  How much energy is required to remove the dielectric?



(a)  C = k Co

     C = (5.6) (20 uF)

     C = 112 uF



(b)  V = Vo / k

 	 V = (10 V) / (5.6)

	 V = 1.8 V



(c)  Uo = 1/2 Co Vo2

	 Uo = 0.5 * 20 uF * (10 V)2

	 Uo = 1000 uJ



	 Uf = 1/2 Cf Vf2

	 Uf = 0.5 * 112 uF * (1.8 V)2

	 Uf = 181 uJ



	 U = Uf - Uo

	 U = 181 - 1000

	 U = 819 uJ