Can current flow in electric shock? Yes. Transient current can flow in insulating medium in the form of time varying displacement current. This was an important discovery made by Maxwell who could predict that electromagnetic waves can be propagated in vacuum.
Note 4: Capacitance and Dielectrics Two conductor plates of area Aseparated by distance d constitute a capacitor. Capacitors are able to store electric energy. (Batteries store chemical energy that can be converted to electric energy.) The electric field due to a surface charge on conductor surafce was found using Gauss' Law E σ Q = = ε ε A ( σ surface charge density) V Since E = (potentail gradient), we find d εa Q εa C Q = V = CV, C = = Capacitance =Farad d V d V d (m) Dimensions of ε : Farad = Farad/m A ( m ) ε means capacitance per unit length in vacuum. V (volts) + + E - - +Q + + + + - - - - -Q
Various Types of Capacitors
A d is small. Dielectric materials (insulators) have permittivities larger than 1 Capacitance C = ε is small quantity because ε = 8.85 1 F/m ε and the capacitance can be increased by filling the capacitor with a dielectric material. (Stud finder detects the chnge in the cpacitance.) If the sepration distance can be made small, a large capacitance can be realized. In electrolytic capacitor, thin layer of oxide insulator is formed on one conductor surface. Electrolytic capacitors have relatively large capacitance but operate at low voltage with fixed polarity. Oil capacitors are used for storing large amount of electric energy for discharging in short time. (The capacitor for the STOR-M tokamak in the Plasma Physics Laboratory can store energy of 3 kj at 7 kv.)
Example 1. Cylindrical capacitor of length l, inner and outer conductor radii a and b The electric field was found earlier, E r 1 Q 1 = πε l r The potential difference between the outer and inner conductors is a 1 Q a1 1 Q b V = Erdr = dr ln b πε l = r πε l a C = = V b a ( ) b We have assumed that the outer conductor is grounded Q πε l ln / ( ) (F) If the capacitor is filled with dielectric material with a permittiv ity ε, Q πεl C = = V ln b / a ( F) ( V = ). The capacitance is
Example : Coaxial cable has inner conductor radius of 1 mm and outer conductor radius of 3 mm. It is filled with Teflon which has a a permittivity ε= ε. Find the capacitance per unit length of the cable. Sol. C / l is C l ( b a) π ε π 8.85 1 ln / ln 3 1 11 = = F/m = 5.6 1 F/m Example 3: Self-capacitance of conducting sphere If a conducting sphere of radius a is charged with a charge Q, the potential of the shpere is raised to V 1 Q = 4πε a as found in Note 3. This defines the self-capacitance of the conducting sphere Q C = = 4 πεa (F) V
Spherical Capacitor Example 4. Spherical capacitor consists of inner spherical electrode of radius a and outer spherical shell of radius b. Find the capacitance. Sol. Let the inner conductor carry a charge Q and the outer conductor Q. From Gauss law, the electric field an be found as 1 Q Er = in the region between the electrodes a < r < b 4πε r The potential difference between the electrodes is a Q a 1 Q 1 Q 1 1 V = Erdr = dr = b 4πε b = r 4πε r b 4πε a b ab The capacitance is C = 4 πε (F) b a a
Parallel and Series Connection of Capacitors Parallel connection The voltage is common. The charges are Q = CV, Q = CV 1 1 The total charge is ( ) Q = Q + Q = C + C V 1 1 Equivalent capacitance is C = C + C eq 1 (parallel connection)
Series connection The charge is common but the voltages differ. V Q =, V = Q 1 C1 C 1 1 = + = + Q 1 1 1 1 Ceq = = or = + V 1 1 + Ceq C1 C C C V V1 V Q C 1 C 1
Example 5. Find the effective capacitance. Sol. Use parallel and series formulae repeatedly.
Energy Storage in Capacitor q A capacitor C having a charge q is at a voltage V =. If the charge is increaesd C by dq, the change in the energy is 1 du = Vdq = qdq C The energy stored in the capactitor is given by 1 Q 1 Q 1 1 U = qdq = = C CV = QV (J) C V du =Vdq area = VQ/ (J) dq Q
Energy Density Example 6. Energy storage in a parallel plate capacitor with plate area A, separation distance d. If it is charged to charge Q, the stored energy is σ 1 Q 1 Q 1 1 U = Ad E Ad C = ε εa/ d = ε = Therefore, an energy density 1 3 u = εe (J/ m ) is associated with electric field E (V/m). Volume ( )
Example 6 : In the arrangement, find the energy stored in each capacitor. Sol. The voltage in each capacitor is: 6 4 3 µ F and µ F: V = 9 V = 9 V = 6 V 3+ 6 + 4 3 6 µ F and 4 µ F: V ' = 9 = 9 = 3 V = 9 6 V 9 6 Energy stored in each capacitor 1 1 CV = ( ) = 1 ( ) = 1 6 3 F: 3 1 6 5.4 mj µ 6 6 F: 6 1 3.7 mj µ ( ) 6 F: 1 6 3.6 mj µ 1 ( ) 6 4 F: 4 1 3 1.8 mj µ = =
Example 7. Before the switches are closed, the capacitor C has a charge Q = CV and C i i i 1 1i 1 i has charge Q = CV. The initial voltage V is common. Find the energy before and after the switches are closed. Sol. The total charge is conserved. C C C C V = C + C V V = V ( ) ( ) 1 1 i 1 f f i C1 + C ( C C ) 1 1 The final energy is U = C V = V 1 f eq f i C1 + C 1 The initial energy is U = ( C + C ) V U The difference is dissipated or radiated. i 1 i f
Dielectric Properties In insulating materials, the permittivity ε is larger than ε. The ratio κ = ε / ε is called the dielectric constant (dimensionless). In typical glass, κ is about.3. Air has κ = 1.54, water has κ = 8 (dc). The origin of higher permittivity in dielectric marerials is due to polarization of atoms and molecules. In dielectrics, many dipoles align themselves along the electric field. If the cpacitor (metal) plate carries a surface charge density σ, the surface charge induced on the dielectric surface is σ induced ( 1 ) = κ σ ( κ ) E which reduces the electric field to E = 1 dielectric
Example 8. Problem 4. Find the dielectric constant of the material inserted. Ans. If the insulator fills the capacitor, the dielectric constant is 85 κ = = 3.4 (Nylon) 5 If it fills partially with area A A' A A' C = ε + κε d d and the voltage becomes Q V = = C ε A κ 1 ε A' + κ ( ) A', the capacitance is
Example 9 A= d =. Problem 47. Capacitor with 5 cm and 1.5 cm is charged to V = 5V and disconnected from the source. The capacitor is then immersed in water which has κ = 8. Determine the capacitor (a) charge, (b) voltage, and energy before and after immersion. (Never do this experiment. Water is not good insulators.) Sol. The capacitance in air and water are C C a w 4 A 1 5 1 1 = ε =8.85 1 =1.475 1 F d 1.5 1 1 = κc = 1.18 1 F a 1 The charge Q= CV a = 3.69 1 C remians unchanged because the capacitor is isolated. The voltage in water is reduced by factor κ, 5 Vw = = 3.15 V 8 1 1 Q 1 (3.69 1 ) 8 Energy in air is Ua = = = 4.61 1 J 1 C 1.475 1 a 1 Energy is water is U w = 5.76 1 J. The difference in energy is dissipated.
Dipole in Electric Field Dipole moment (vector) p = qd where d is the distance between q and q. If palced in an electric field E, the dipole is subject to torque τ = pe sinθ In vector form, τ = E p (N m) The positive charge is at a potential which is lower than than the potential of negative charge. Therefore, the potential energy of a dipole in electric field is U = qde cos θ = p E (J) E +q qe d θ -qe -q
Example 8. Capacitor partially filled with a dielectric A fraction f of capacitor (area A, plate distance d) is filled with dielectric having dielectric constant κ. Find the capacitance. Sol. C C eff = ε 1 A ( 1 f ) 1 = = 1, C d = κε ( 1 ) The electric field in the dielectric slab is smaller than that in air capacitor by a factor κ. The voltage is divided as C C V = V, V = V 1 1 C1 + C C1 + C A fd CC κ ε A C + C f + κ f d