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Ch 26.4 – Energy Stored in a Capacitor – charging a capacitor Switch goes shut. Battery establishes E-field in wires, charge builds up on the capacitor. Chemical energy in battery goes down and electrical potential energy stored in capacitor goes up. Electrical potential energy results from charge separation across capacitor’s plates Ch 26.4 – Energy Stored in a Capacitor – charging a capacitor To charge the capacitor, an external agent (battery) must do work to separate the charges. Step by step, the battery “grabs” a small amount of charge dq off one capacitor plate and move it to the other. At first, this requires no work, because the uncharged capacitor has no electric field to resist the movement of charge. But, once dq has been transferred, the capacitor starts to develop a potential difference. Now, to move more charge from one plate to the other, the battery must do some amount of work “dW,” to overcome the rising potential difference across the plates. As more and more charge is transferred, the work required to transfer the same amount of charge, dq, goes up. Ch 26.4 – Energy Stored in a Capacitor – charging a capacitor Suppose q is the amount of charge on the capacitor at some instant during the charging process. At the same instant, the potential difference across the capacitor is ΔV = q/C The amount of work necessary to transfer dq amount of charge across this potential difference is dW = ΔVdq = (q/C)dq Ch 26.4 – Energy Stored in a Capacitor Thus, the total amount of work required to charge the capacitor from q = 0 to a final charge of q = Q is Q W q C 0 dq 1 C Q 0 qdq Q 2 2C But, in an isolated system with no non-conservative forces, total mechanical energy must be conserved. Therefore, the work done to charge the capacitor must equal the change in the system’s potential energy. Ch 26.4 – Energy Stored in a Capacitor U Q 2 2C 1 2 QV 1 2 Energy stored in a charged capacitor C V 2 Ch 26.4 – Energy Stored in a Capacitor U Q 2 2C 1 2 QV 1 2 C V 2 -It’s not obvious, but the potential energy stored in the capacitor actually resides in its electric field. -This implies we should be able to solve the density of the energy stored in the field (J/m3). -For a parallel plate capacitor, we already know: -and, its capacitance is just: C V Ed 0A d -Substituting these into the purple equation, U 1 0A 2 d E 2 d 2 1 2 0 Ad E 2 -Dividing by the volume in between the plates of the capacitor (V=Ad), we get Energy per unit volume in 1 2 uE 0E a capacitor (J/m3) 2 Ch 26.4 – Energy Stored in a Capacitor -We don’t attempt it here, but it can be shown that this result is valid for any electric field! uE 1 2 0E 2 Energy per unit volume in an electric field. -In a very real sense, electric fields “carry” energy. EG 26.4 – Rewiring two Charged Capacitors Two capacitors, C1 and C2 (C1 > C2), are charged to the same initial potential difference, ΔVi. The charged capacitors are removed from the battery, and their plates are connected with opposite polarity, as shown. The switches, S1 and S2, are then closed. (a) Find the final potential difference ΔVf between a and b after the switches are closed. (b) Find the total energy stored in the capacitors before and after the switches are closed and determine the ratio of the final energy to the initial energy. Ch 26.5 – Capacitors with dielectrics A dielectric is something you stick in between the plates of a capacitor to change (increase) it’s capacitance. The term comes from the fact that, at the atomic level, such materials can be polarized into arrays of dipoles. Ch 26.5 – Capacitors with dielectrics Common dielectric materials are: wax, paper, oil, polymers, fluid (electrolyte), etc. Dielectrics are insulators. Since you stick the dielectric material into the region of the capacitor’s E-field, it changes how “good” the capacitor is at establishing the E-field ΔV↓ for the same amount of Q on the capacitor plate. What happens? Ch 26.5 – Capacitors with dielectrics Consider parallel-plate capacitor where ΔV0 = Q0/C0 Assume no battery is connected Q can’t change When you stick a dielectric in between the plates Ch 26.5 – Capacitors with dielectrics Consider parallel-plate capacitor where ΔV0 = Q0/C0 Assume no battery is connected Q can’t change When you stick a dielectric in between the plates -where κ is a dimensionless constant called the “dielectric constant” V V0 Ch 26.5 – Capacitors with dielectrics -Q on the capacitor does not change -Therefore: C Q0 V Q0 V0 C C 0 -the capacitance is changed by a factor of κ. -as κ goes up, C goes up. Q0 V0 Ch 26.5 – Capacitors with dielectrics -For a parallel plate capacitor C0 0A d C 0A d To make capacitance ↑ -decrease d -increase A -increase κ - Only limited by “dielectric strength” of the dielectric Example values of dielectric constant “Dielectric strength” is the maximum field in the dielectric before breakdown. (a spark or flow of charge) E max Vmax / d EG 26.5 – Energy stored before and after A parallel-plate capacitor is charged with a battery to a charge of Q0. The battery is then removed, and a slab of material that has a dielectric constant κ is inserted between the plates. Identify the system as the capacitor and the dielectric. Find the energy stored in the system before and after the dielectric is inserted. EG 26.4 – Rewiring two Charged Capacitors A parallel-plate capacitor is charged with a battery to a charge of Q0. The battery is then removed, and a slab of material that has a dielectric constant κ is inserted between the plates. Identify the system as the capacitor and the dielectric. Find the energy stored in the system before and after the dielectric is inserted. 2 Before: Q0 U0 2C 0 2 After: U0 Q0 2C 2 U0 Q0 2 C 0 Where did the energy go? U0 Ch 26.6 – Electric Dipole in an Electric Field The combination of two equal charges of opposite sign, +q and –q, separated by a distance 2a Every dipole can be characterized by it’s “dipole moment.” - vector which points from –q to +q -magnitude p = 2aq p1 p2 p p1 p 2 Ch 26.6 – Electric Dipole in an Electric Field What happens when we pop this baby in an external E-field? Ch 26.6 – Electric Dipole in an Electric Field What happens when we pop this baby in an external E-field? -external field exerts F=qE on each charge -net torque about the dipole’s center -dipole rotates to “align” with the field Ch 26.6 – Electric Dipole in an Electric Field What happens when we pop this baby in an external E-field? -external field exerts F=qE on each charge -net torque about the dipole’s center -dipole rotates to “align” with the field Fa sin Fa sin Ch 26.6 – Electric Dipole in an Electric Field net 2 Fa sin but, F qE and p 2 aq Thus: 2 aqE sin pE sin Ch 26.6 – Electric Dipole in an Electric Field net 2 Fa sin but, F qE and p 2 aq Thus: 2 aqE sin pE sin pE Ch 26.6 – Electric Dipole in an Electric Field The dipole and the external field are a system -electric force is an internal conservative force we can describe its work using a potential energy In other words, different configurations of the dipole-field system have different potential energies. Ch 26.6 – Electric Dipole in an Electric Field As the dipole aligns with the field, the system’s potential energy goes down. Ch 26.6 – Electric Dipole in an Electric Field -Work must be done to “un-align” the dipole from the field. -in an isolated system, the work input must correspond to an increase in potential energy. Ch 26.6 – Electric Dipole in an Electric Field -Work must be done to “un-align” the dipole from the field. -in an isolated system, the work input must correspond to an increase in potential energy. W = ΔK + ΔU Ch 26.6 – Electric Dipole in an Electric Field -To rotate the dipole through some small angle dθ, an amount dW of work must be done. dW d but, pE sin Ch 26.6 – Electric Dipole in an Electric Field -To rotate the dipole through some small angle dθ, an amount dW of work must be done. dW d but, pE sin -so, to rotate the dipole from θi to θf, the change in potential energy is: U f Ui f d pE sin d i f i pE f sin d i pE [ cos ] if pE (cos i cos f ) Ch 26.6 – Electric Dipole in an Electric Field Let’s define the zero potential energy as being when the dipole is at θ = 90, Ui 0 when i 90 Ch 26.6 – Electric Dipole in an Electric Field Let’s define the zero potential energy as being when the dipole is at θ = 90, Ui 0 i 90 when We’ll use this reference energy as an anchor point. At any time, we can write the system’s instantaneous potential energy, U, with respect to the zero-point potential energy. U U f Ui U f U 90 U f 0 Ch 26.6 – Electric Dipole in an Electric Field Let’s define the zero potential energy as being when the dipole is at θ = 90, Ui 0 i 90 when We’ll use this reference energy as an anchor point. At any time, we can write the system’s instantaneous potential energy, U, with respect to the zero-point potential energy. U U f Ui U f U 90 U f 0 But, we already know pE [ cos ] if pE (cos i cos f ) pE cos f U pE cos Ch 26.6 – Electric Dipole in an Electric Field Let’s define the zero potential energy as being when the dipole is at θ = 90, Ui 0 i 90 when We’ll use this reference energy as an anchor point. At any time, we can write the system’s instantaneous potential energy, U, with respect to the zero-point potential energy. U U f Ui U f U 90 U f 0 But, we already know pE [ cos ] if pE (cos i cos f ) pE cos f U pE cos U pE EG 26.6 – The Water Molecule A water molecule has an electric dipole moment of 6.3x10-30 Cm. A sample contains 1021 water molecules. All of the dipoles are oriented in the direction of an external E-field, which has a magnitude of 2.5x105 N/C. How much work is required to rotate all the dipoles from this orientation (θ = 0) to one in all the dipoles are perpendicular to the external field (θ = 90)? EG 26.6 – The Water Molecule A water molecule has an electric dipole moment of 6.3x10-30 Cm. A sample contains 1021 water molecules. All of the dipoles are oriented in the direction of an external E-field, which has a magnitude of 2.5x105 N/C. How much work is required to rotate all the dipoles from this orientation (θ = 0) to one in all the dipoles are perpendicular to the external field (θ = 90)? U W W U 90 U 0 ( NpE cos 90 ) ( NpE cos 0 ) NpE (10 1 . 6 10 3 J 21 )( 6 . 3 10 30 C m )( 2 . 5 10 N / C ) 5 Example P26.9 When a potential difference of 150 V is applied to the plates of a parallelplate capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the spacing between the plates? Q d 0 V 0 A 8.85 10 30.0 10 9 V d 12 C cm C 2 2 N m 2 150 V 1.00 10 4 cm 2 m 2 4.42 m Example P26.21 Four capacitors are connected as shown in Figure P26.21. (a) Find the equivalent capacitance between points a and b. (b) Calculate the charge on each capacitor if ΔVab = 15.0 V. 1 Cs 1 15.0 1 Q C V 5.96 F 15.0 V 3.00 V C s 2.50 F C eq C 89.5 C 20.0 F 4.47 V 15.0 4.47 10.53 V C p 2.50 6.00 8.50 F 1 1 8.50 F 20.0 F Q 89.5 C Q C V 6.00 F 10.53 V 1 5.96 F 89.5 63.2 26.3 C 63.2 C on 6.00 F Example P26.27 Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C1 = 5.00 μF, C2 = 10.0 μF, and C3 = 2.00 μF. 1 1 Cs 5.00 10.0 1 3.33 F C p1 2 3.33 2.00 8.66 F C p 2 2 10.0 20.0 F C eq 1 1 8.66 20.0 1 6.04 F Example P26.35 A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled? d2 2d1 , C2 1 2 C1 . Therefore, the stored energy d ou bles Example P26.43 Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm2 and plate separation of 0.040 0 mm. C 0 A V m d ax 2.10 8.85 10 Em 12 F m 4.00 10 ax d 1.75 10 5 6 60.0 10 m V m 4 m 2 8.13 10 4.00 10 5 m 11 F 81.3 pF 2.40 kV Example P26.59 A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00 × 108 V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4 000 V. Find the minimum area of the capacitor plates. 3.00 Em C A Cd 0 C V m ax 2.00 10 0 A d 8 0.250 10 6 V m V m ax d F 0.250 10 4 000 3.00 8.85 10 2.00 10 6 ax 0 E m ax 12 8 0.188 m 2