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ELECTRIC POTENTIAL ENERGY AND ELECTRIC POTENTIAL • POTENTIAL ENERGY • ELECTRIC POTENTIAL • WORK-ENERGY THEOREM • CAPACITANCE • COMBINATIONS OF CAPACITORS • STORED ENERGY Written by Dr. John K. Dayton POTENTIAL ENERGY IN A UNIFORM ELECTRIC FIELD: WAB F cos x F qE WAB qE x U AB WAB U AB qE x In this example a charged particle is moved from point A to point B in a uniform field by the electrostatic force. The work done by the force and the change in potential energy of the particle can be calculated in the usual way. Remember, energy is a scalar quantity. ELECTRIC POTENTIAL: Electric potential is defined as the electric potential energy per unit charge. V U q V U q In a uniform electric field, where U = -qE x, the change in electric potential will be V = -Ex. The SI unit for electric potential is the volt, V. V=J/C. To find a particle’s change in potential energy, use: U q V Electric potential is a scalar quantity. EXAMPLE: What is the change in a particle’s potential energy if it moves from a position of 100 volts to a position of 150 volts? Let the particle have an electric charge of -3mC. Click For Answer U qV U 3mC 150V 100V U 150m J THE ELECTRIC POTENTIAL OF A POINT CHARGE: q V k r VA B VB VA VA B q q k k rB rA VA B 1 1 kq rB rA If q is a point charge (positive or negative), then the electric potential a distance r from q is given by the top equation. Note, V will be negative if q is negative. The difference between the potentials at two points is also shown. r V V vs r for a positive charge V vs r for a negative charge r V THE SUPERPOSITION OF ELECTRIC POTENTIALS OF SEVERAL POINT CHARGES: Let P be a point in space near several point charges such that it is a distance r1 from q1, r2 from q2, and r3 from q3. Then the net electric potential at P is: VP VP ,1 VP ,2 VP ,3 q1 q2 q3 VP k k k r1 r2 r3 EXAMPLE: Calculate the electric potential on the x axis at 15.0 cm produced by two point charges; q1 = -4.0 mC on the origin and q2 = +3.0 mC at 10.0cm on the y axis. Click For Answer q2 = +3mC + r2 = .1803m 10cm - r1 = .15m 15cm r2 .1m .15m 2 2 0.1803m VP VP ,1 VP ,2 P This diagram is the first step in a well planned solution. q1 = -4mC Solution continues on next slide. continued from previous slide… kq1 kq2 VP r1 r2 VP 8.99 10 Superposition of point charge potentials 9 Nm 2 C2 4 10 C .15m VP 9.03 104V 6 8.99 10 9 Nm 2 C2 6 3 10 C .1803m Electric potential is a scalar so is much easier to work with than the electric field. No directions are involved. Don’t forget to use the sign of the charge when calculating an electric potential. THE WORK-ENERGY THEOREM FOR ELECTROSTATICS: A particle of charge q and mass m moves under the influence of an electric field from point A to point B. K U 0 1 2 In General: mv B2 12 mv A2 q V A B 0 V A B V B V A For a single point charge: V A B For a uniform field E: V A B For an infinite plane of charge, E 2òo : For a charged, infinite conducting sheet, E : òo 1 1 kqo rB rA E x B bA V A B xB x A 2 òo V A B xB x A òo EXAMPLE: A proton is released from rest 5.0 cm from the surface of a charged sphere of radius 10.0 cm and charge Q = 4.0 mC. What will the proton’s speed be when it has moved 1 meter? The proton will move between ri = .15m and rf = 1.15m. Work-Energy Theorem for 0 K f Ki q V f Vi Click For Answer Electrostatics kQ kQ Working Equation with 2 1 0 2 mv f 0 e Potentials of a spherical charge r ri f 2ekQ 1 1 vf m ri rf vf 2 1.6 10 v f 7.21 106 ms 19 Solved for final velocity C 8.99 10 9 Nm 2 C2 1.67 1027 kg Final Answer 4 10 C 6 1 1 .15m 1.15m CAPACITANCE AND THE CAPACITOR: A capacitor is comprised of two charged, conducting bodies maintained at a potential difference. The charge in the capacitor is actually a charge separation. Capacitance is defined as the ratio charge to potential difference within a capacitor. Q C V C = capacitance in SI units of coulombs/ volts. This combination is called the farad, F. Q = charge on capacitor. One conductor has +Q while the other has -Q. V = voltage difference between conductors. A capacitor’s capacitance depends on its size and shape, not on the charge separation and not on the voltage difference. THE PARALLEL PLATE CAPACITOR: The parallel plate capacitor is comprised of two metal plates that face each other with each plate connected to a battery terminal. The inside area of each plate is A and they are separated by a distance d. d A E +Q V -Q The plate connected to the positive battery terminal will have a charge of +Q on its inside surface. The plate connected to the negative terminal will have -Q on its inside surface. The voltage difference between the plates will be the battery voltage, V. The electric field between the plate will be uniform given by E / òo pointing from the positive plate to the negative plate. Q A E òo A V Ed Q E òo A C V Ed òo A C d EXAMPLE: Calculate the capacitance of a parallel plate capacitor made of two circular disks of radius 6.0 cm and separated by 0.5mm. How large a radius should they have if the capacitor is to be 1.0 F? (a) òo A òo r 2 C d d 8.85 10 C Click For Answer (b) .06m 12 C 2 Nm 2 3 2 .5 10 m C 2.00 1010 F 200 pF òo A òo r 2 C d d r dC r òo 3 .5 10 m 1F 8.85 10 12 C 2 Nm 2 r 4.24 103 m CAPACITORS CONNECTED IN SERIES: C1 C2 + V Ceq + - V - Beginning with a group of capacitors connected in series, find the single, equivalent capacitor. Q Q V1 V2 C1 C2 In series each capacitor has the same charge on it: Q1 = Q2 = Qeq = Q. Q Q Q Ceq C1 C2 In series the voltages across each capacitor add to the battery voltage: V1 + V2 = Veq = V. 1 1 1 Ceq C1 C2 V V1 V2 Q V Ceq EXAMPLE: C1 = 4.0 mF and C2 = 6.0 mF are connected in series to a 24V battery. What is the stored charge in C1? C1 C2 + V Ceq Click For Answer - 1 1 1 1 1 Ceq C1 C2 4 m F 6m F + V Use series equation Ceq 2.4 m F Charge stored in Qeq CeqV 2.4 m F 24V 57.6mC C . eq Capacitors in series each have the Q1 Q2 57.6mC same charge. CAPACITORS CONNECTED IN PARALLEL: C1 + + - Ceq V C2 Beginning with a group of capacitors connected in parallel, find the single, equivalent capacitor. In parallel each capacitor has the same voltage across it: V1 = V2 = Veq = V. The individual charges add to the charge on the equivalent capacitor: Q1 + Q2 = Qeq. - V Q1 VC1 Q2 VC2 Qeq VCeq Qeq Q1 Q2 VCeq VC1 VC2 Ceq C1 C2 EXAMPLE: C1 = 4.0 mF and C2 = 6.0 mF are connected in parallel to a 24.0 V battery. What is the stored charge in C1? C1 Ceq + + - V C2 V Click For Answer - Ceq C1 C2 4 m F 6m F 10m F Equation for parallel Qeq CeqV 10 m F 24V 240mC Charge in Equivalent Capacitor V1 V2 Veq 24V Capacitors in parallel have the same voltage Q1 C1V1 4 m F 24V 96mC CAPACITORS CONNECTED IN GENERAL: In Parallel: In Series: Ceq Ci 1 1 Ceq Ci C3 In the diagram C1 and C2 are not in series; they cannot be directly combined. C2 and C3 are in parallel and can be combined. Once C2,3 is known it can be combined with C1 in series. C1 C2 + V - ENERGY STORED IN A CAPACITOR: Capacitors store energy within their electric fields. Assume the average voltage during the charging process is one-half the final voltage, V/2. Thus the charge that separates crosses this voltage and the net change in potential energy is QV/2. This is the energy stored in the capacitor: U 12 QV 12 CV 2 Calculate the energy density within a parallel plate capacitor: òo A C d Vol Ad V Ed The final expression is good for any capacitor. 1 2 Q2 C 2 U CV u 12 Vol Vol 2 òo A Ed d u 12 Ad u 12 òo E 2 EXAMPLE: Calculate the equivalent capacitance of the circuit and the energy stored in each of the original capacitors. C1 C2,3 C3 = 2mF C1 = 2mF + + Click For Answer C2 = 3mF - Ceq V - V + V A series of reduced circuits. C2 and C3 can be combined first because they are in parallel. Solution Continues on Next Slide continued from previous slide… C2,3 C2 C3 3m F 2 m F 5m F 1 1 1 1 1 Ceq C1 C2.3 2 m F 5m F Compute equivalent capacitor Ceq 1.4286m F Qeq CeqVBat 1.4286m F 24V 34.2857 mC Q1 Q2,3 34.2857 mC (Because they are in series) Q1 34.2857 mC V1 17.1429V C1 2m F Q2.3 34.2857 mC V2,3 6.8571V C2.3 5m F Distribute charge and voltage up one circuit level continued from previous slide… V2 V3 V2.3 6.8571V (Because they are in parallel) Q2 C2V2 3m F 6.8571V 20.5713mC Q3 C3V3 2 m F 6.8571V 13.7142 mC 2 m F 17.1429V 293.879 m J 2 2 1 1 U 2 2 C2V2 2 3m F 6.8571V 70.5297 m J 2 2 1 1 U 3 2 C3V3 2 2 m F 6.8571V 47.0198m J U1 C V 2 1 1 1 2 2 1 2 total 411.429 m J U eq C V 1 2 2 eq Bat 1 2 1.4286m F 24V 2 A N S W E R S 411.429 m J The total of the individual stored energies must equal the energy stored in the equivalent capacitor. End Of Presentation