C = V Q. To find the capacitance of two conductors:

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1 Capacitance

2 Capacitance is a measure of the ability of two conductors to store charge when a given potential difference is established between them. Two conductors, on one of which is charge +Q and on the other Q. The capacitance of the conductors is defines as C = Capacitance is measured in farads (F). Note that the capacitance is a proportional factor between the charge and the potential difference between the conductors. Q V To find the capacitance of two conductors: First: Place charge +Q on one conductor an Q on the other. Second: Measure the potential difference between the conductors Third: Since V and Q are related by where the proportionality constant is /C V Q The farad is the unit of capacitance. microfarad (μf) = 0-6 farad (F) picofarad (pf) = 0 - farad (F)

3 Capacitance of two parallel plates When the plates of a capacitor are charged with a charge Q (+Q and Q) on opposite plates), it can be shown that there is a uniform electric field between the plates, which has a magnitude E Two parallel plates of area A are separated a distance d Thus VV = EEEE = QQQQ εε 0 AA parallel-plate capacitor CC = QQ VV = AAεε oo dd The capacitance of the parallel plate capacitor is directly proportional to the plate area and inversely proportional to the separation, with a constant added to make the units work out correctly. The capacity is entirely geometry dependent.

4 Capacitors as lumped circuit Elements Capacitors are used in electric circuits, so it is necessary to know how capacitors add when used in series and parallel. First look at three capacitors placed in series. The applied voltage is distributed between the three capacitors so V=V +V +V 3, since there is no way to add charge except to the end plates, those connected directly to the battery. The equation for the voltage is: V=V +V +V 3, so Therefore C 3 Q Q V = V + V + V3 = + + C C = + + can be replaced by an equivalent capacitance C C C Q C 3 C T Capacitors placed in series adds in a reciprocal manner: CC TT = CC + CC + CC 3

5 Parallel arrangements Capacitor show in the figure is connected in parallel across a voltage V. Both capacitors are at the same voltage, so the charge distribution between both capacitors must be equal to the total Q provided by the battery: By using the definition of capacitance V=V =V =V 3 C = Q V + CV + C3V = V ( C + C C3 Q = C V + we have: ) so can be replaced by an equivalent capacitance; Capacitors placed in parallel ads to give the total capacitance: C = ( C + C + ) 3 C

6 Spherical capacitor is compose of two concentric spheres Φ = EE dddd = qq εε 0 qq = ρρρρrr LL where, EE 4ππRR = qq εε 0 The potential difference between the plates of a capacitor is related to the field E V b V a b = E dr a q Vb Va = dr 4πR Which the integral is evaluated along the path that starts on one plate and ends on the other. However for a capacitor with separation between plates A spherical capacitor is composed of two concentric metal spherical shells, of radii a and b. As a Gaussian surface we draw a sphere of radius r concentric then there is an electric field between the two surfaces of radii a and b (a < b). The difference of potential between a and b is: b a V b V a = + q q dr q dr = = πε0r 4πε 0 r 4πε a 4 b 0 a b V = q 4πε 0 b a ab CC = qq VV = 4ππεε ooaaaa bb aa spherical capacitor

7 A cylindrical conductor is charged wit constant density ρ. Compute the electric field at r < R and r > R EE dddd = QQ εε 0 where, QQ = ρρ ππrr LL EE dddd = EE dddd dddd = EEEEEEEEE QQ = ρρ dddd dddd = ρρ ππrr LL εε 0 εε 0 rr < RR EEEEEEEEE = ρρ ππrr LL εε 0 EEEEEEEEE = ρρ ππrr LL εε 0 ρρ ππ EE = rr εε 0 rr > RR EEEEEEEEE = ρρ ππrr LL εε 0 EE = ρρ ππrr εε 0 rr

8 A cylindrical capacitor is composed of two concentric metal cylindrical shells, of radii a and b, and charge density ρ C/m 3. As a Gaussian surface we draw a cylinders of radius r concentric, there is an electric field between the two surfaces of radii a and b (a < b). The electric field between a and b is: EE dddd = qq εε 0 where, qq = ρρρρrr LL EEEππππππππ = EEEππrrLL = qq εε 0 EEππππππ = ρρρρrr LL εε 0 EE = ρρrr εε 0 rr rr < RR EE = ρρrr rr = RR EE = ρρrr rr > RR EE = ρρrr εε 0 εε 0 εε 0 rr The difference of potential between a and b is: VV = aa bb EE dddd = ρρrr εε 0 aa bb rr dddd VV = ρρrr εε 0 ln bb/aa Capacitance: CC = qq VV = ρρρρrr LL ρρrr εε ln bb/aa 0 = εε 0 ππππ ln bb/aa Capacitance per length unit: CC LL = ππεε 0 ln bb/aa

9 , Energy stored in a capacitor Work is required to charge a capacitor. Hence, a charged capacitor has electric potential energy. As the charge on the plates goes from zero to a total charge Q in the charging process, the potential difference or voltage across the plates goes from zero to V. Work is done in moving charge (from the negative plate to the positive plate) through the increasing potential difference. To compute this work, we use an average potential difference V V Vfinal Vinitial V 0 = = = V V initial = 0.Then he work or, equivalently, the potential energy of the charged capacitor is ( V) W U QV Q = = = U = QV Q = CV We refer to this as the energy stored in a charged capacitor. Using the defining equation for capacitance, the potential energy can be written in three forms: Q U = QV = = C CV The last form is usually the most convenient in practical applications.

10 Using a parallel plate capacitor as a model observes experimentally that the amount of charge per unit voltage increases with the insertion of various dielectric materials. The easiest way to explain this is to say that C=Aε o /d is changed by a factor k, so C=kAε o /d with this k called the dielectric constant, a measure of the amount by which the capacity of a capacitor is increased due to a particular dielectric. The mechanism of increasing the capacitance can be thought of as the dipoles lining up across the separation d with the positive end of the string effectively neutralizing some of the negative charge on the plate and the negative end of the string doing the same to the other plates. Consider another experiment with dielectrics. Connect a battery in parallel with identical sized parallel plate capacitors except one has a dielectric and the other does not. C = A / d C = k A / d ε 0 Since, q=c/v to carry the same charge the capacitor with the dielectric would have to have V o likewise E o./k Finally looking at the energy density ε 0 reduced to V o /k and u = CV Ad = kε 0 AV Ad = kε 0 E

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