5C20.10 Parallel Plate Capacitor Dielectrics Abstract A voltage is placed across a parallel plate capacitor by using the electrophorus to put a charge on one side of the capacitor. The presence of the charge can be seen on the electroscope. Once charged, a dielectric material can be placed between the plates of the capacitor. The electric field between the two plates causes an induced electric field of opposite polarity across the dielectric. The net electric field is reduced causing a decrease in the voltage across the capacitor. This change in the voltage can be detected by the electroscope. Picture Setup Setup is 5 minutes Safety Concerns None. Equipment • Dielectric apparatus • Fur • Transparent electroscope • Electrophorus • Circular metal plate 1 Procedure Connect one of the parallel plates to the black ring on the transparent electroscope and the other to the outer ring of the electroscope. Adjust the gap between the plates so that the dielectric can be easily inserted into the gap. Remove dielectric. Rub the electrophorus with the fur, place the circular metal plate on the electrophorus and press down on it with a finger. There should be a slight electrical discharge. Pickup the circular plate and touch it to the top plate of the capacitor at which time there should be another electrical discharge and the needle of the electroscope should have deflected slightly from center. Repeat charging the upper plate until the needle of the electroscope is near its maximum displacement. Once charged return the dielectric into the gap between the parallel plates at which time the electroscope should show a decrease in the voltage across the capacitor. On humid days the electrophorus may not work very well. If this happens try using the Wimshurst generator to provide the required voltage. Attach the sides of the Whimshurst’s leyden jar capacitor different sides of the parallel plate capacitor using the leads and the alligator clips. Spin the generator crank in the clockwise direction to charge up the plates. When finished use the discharging rod to discharge the plates. When complete, place the dielectric between the plates and adjust the vertical positions of the plates so that they are as close to the dielectric as possible while still allowing the dielectric to be easily removed. Rotate the plates so that are aligned with each other while still covering the entire surface of the dielectric. Stack electrophorus and circular disk on base of dielectric apparatus and place in cupboard. Return fur to box of furs, and return electroscope to the cupboard with the rest of the electroscopes. Theory The electric field between the two parallel plates of a capacitor, which have equal and opposite charge densities of σ, is illustrated in Figure 1a and is described by the equation, Eo = σ , o (1) where Eo is the electric field between two parallel plates with no dielectric, and o is permittivity of free space (o = 8.85x10 − 12C 2 /N m2 ). When a dielectric is placed between the plates of the capacitor the electric field will +σ −σ induce a charge density of magnitude σi , Eo E – on the sides of the dielectric. A dielectric is – + + an insulator and therefore the charges in it – + – + – + are bound to the atoms or molecules. The charges are not free to move around the – – + + dielectric, they can only be shifted within each atom or molecule. As a result the – – – + + + induced charge density is less than charge density of the plates of the capacitor. – – + + Since the polarity of the induced charge – + – + – + density will be opposite the polarity of the capacitor as shown in Figure 1b, part of – – + + the charge density of the capacitor will be neutralized. A lower net charge density will +σi −σi +σ −σ resulting in a smaller electric field across the capacitor. (a) (b) The electric field across a parallel plate Figure 1: Diagram showing the electric field lines between parallel capacitor with a dielectric is described by, plates without dielectric material in Figure 1b and with dielectric σ − σi material in Figure 1a. The charge densities within the plates, σ and E= , (2) o the induced charge densities within the dielectric material, σi are also labelled. where E is the electric field across the capacitor with the dielectric in place. The 2 ratio of the electric field across a capacitor without any dielectric to the electric field across the same capacitor with a dielectric is a characteristic of the material of which the dielectric was made. This constant is obtained by taking the ratio of Equations 1, and 2. The resulting expression is Eo σ = = K, E σ − σi (3) where K is the dielectric constant. Note that the induced charge density cannot be larger that the original charge density, or rather σi ≤ σ, therefore K ≥ 1. The voltage across a parallel plate capacitor is given by the expression, V = Ed, (4) where d is the distance between the two plates of the capacitor. Substituting Equation 4 into Equation 3 reveals the expression, Vo , (5) V = K where Vo is the initial voltage across the capacitor without the dielectric, and V is the final voltage across the capacitor with the dielectric in place. Equation 5 gives the relationship between the initial and final voltages across the capacitor. Since K ≥ 1 the final voltage must always be less then or equal to the initial voltage. This prediction is confirmed by the reaction of the electroscope when a dielectric is placed between the plates of a charged capacitor. 3 References [1] Freier, G. D. and Anderson, F. J. A Demonstration Handbook for Physics, ”Ed-2 Dielectrics”, American Association of Physics Teachers One Physics Ellipse, College Park MD, 1996. pg E-19. [2] Sutton, Richard Manliffe. Demonstration Experiments in Physics, ”E70. Dielectric Constant and Capacitance”, McGraw-Hill Book Company Inc., New York and London, 1938. pg 277. 4