Where A is the plate area and d is the plate separation.

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Abel Watts
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1 DIELECTRICS Dielectrics an the parallel plate capacitor When a ielectric is place between the plates of a capacitor is larger for the same value of voltage. From the relation C = /V it can be seen that the capacitance must also increase. The ratio of the capacitance of the capacitor with the ielectric to the capacitance of the capacitor without the ielectric is calle the ielectric constant κ of the material. If the same charge is maintaine on the capacitor with an without the ielectric then the potential ifference between the plates of the capacitor with the ielectric, V will be less than that without the ielectric V by a factor of 1/κ. V = V /κ C= V EAε = = E Aε Where A is the plate area an is the plate separation. Therefore: C= A κε Dielectric materials - a escription Dielectrics can be of two types, those which possess permanent ipole moments such as water an those which obtain an inuce ipole. When an electric fiel is applie to these materials the ipoles ten to align themselves to the fiel - this is not a perfect alignment ue to thermal effects. The alignment is improve by either increasing the fiel or by ecreasing the temperature. The alignment is ue to the electric ipole moment p which is proportional to the electric fiel.

2 To illustrate the ipole effects taking place within a slab of ielectric, we can take a charge parallel plate capacitor (battery isconnecte) which has a fixe charge an provies a uniform electric fiel E. When the ielectric is place between the plates then the ipoles align with the electric fiel an the centre of positive charge separates from the centre of negative charge i.e. the ielectric becomes polarise while remaining electrically neutral. This separation of charge is on the atomic scale an it shoul be note that the charge oes not move as it woul if the slab were mae from a conuctor - no charge movement over macroscopic istances. The effect of this charge separation is the introuction of a electric fiel E which opposes the external fiel E, the resultant fiel E is therefore the vector sum of these two fiels: E = E + E which is smaller than the original fiel. From the euation for a parallel plate resistor (V = E) it can be seen that the fiel is irectly proportional to the potential ifference an therefore the reuction in the overall fiel results in a reuction in the potential ifference between the plates, an: E /E = V /V = κ If the battery is left connecte uring the introuction of the ielectric, then the above euation oes not hol. The potential ifference now remains constant but the charge on the plates increases by a factor of κ. The use of Gauss s law for capacitors with a ielectric

3 If no ielectric is present then Gauss s law gives: ε E.A = ε EA = E = D1 ε A With the ielectric present then Gauss s law gives: ε E.A = ε EA = ' E ' = εa εa D Using E = E /κ an substituting in D1 we get: E E = = κ κε A By combining this euation with D we fin: ' = κε A ε A ε A ' = 1 1 κ the surface inuce charge, is shown to be always less than the magnitue of the free charge an is eual to zero when there is no ielectric i.e. κ = 1. Returning to the integral for the case with a ielectric it can be shown that: ε κ E.A = This euation generally hols for all capacitors an is use when a ielectric is present.

4 Energy within a capacitor Work must be one to separate two eual an opposite charges an this energy can be store in the system i.e. on the capacitor plates. The energy can be recovere if the charges are allowe to come back together. If a capacitor is initially uncharge, then the work W one to charge the capacitor is eual to the electric potential energy U store by the charge capacitor. This can be visualise as pulling electrons from one plate an epositing them onto the other plate. If at time t a charge has been transferre from one plate to the other plate of a capacitor, then the potential ifference V will be eual to /C. If now, an extra small amount of charge is transferre, then the extra work neee to o this will be eual to: W = V = ( /C) If this process is continue until the total charge is, then the total work one will be: W ' = W = C ' = 1 C Substituting for using the stanar relation, = CV we obtain: W = U = 1 CV The energy store in a capacitor is sai to resie in the electric fiel. In a parallel plate capacitor, if we neglect fringing at the eges, then the electric fiel has the same value for all points between the plates. Therefore the energy store per unit volume (the energy ensity) is uniform an is given by: U u = = A 1 CV A where A is the area of the plates an is the plate separation: A is therefore the volume If we remember that the capacitance C for a parallel plate capacitor is: A C = κε then by substitution: u = κε V Now the electric fiel E = V/ so:

5 u = 1 κε E This euation was erive for the parallel plate capacitor but it also hols true for all capacitors. In general, if we have an electric file E at any point in space, we can think that at that point there is a site of store energy of magnitue u = 1 κε E per unit volume.

CAPACITANCE: CHAPTER 24. ELECTROSTATIC ENERGY and CAPACITANCE. Capacitance and capacitors Storage of electrical energy. + Example: A charged spherical

CAPACITANCE: CHAPTER 24. ELECTROSTATIC ENERGY and CAPACITANCE. Capacitance and capacitors Storage of electrical energy. + Example: A charged spherical CAPACITANCE: CHAPTER 24 ELECTROSTATIC ENERGY an CAPACITANCE Capacitance an capacitors Storage of electrical energy Energy ensity of an electric fiel Combinations of capacitors In parallel In series Dielectrics