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Question 1 Which one of these statements is true? When near the center of the object, the electric field hardly changes with increasing distance from A. a uniformly charged long rod. B. a uniformly charged ring. C. a uniformly charged disk. D. a point charge. Question A solid metal ball carrying negative excess charge is placed near a uniformly charged plastic ball. -Q +Q plastic metal Which one of the following statements is true? A. The electric field inside both balls is zero. B. The electric field inside the metal ball is zero, but it is nonzero inside the plastic ball. C. The electric fields inside both objects are nonzero and are pointing toward each other. D. The electric field inside the plastic ball is zero, but it is nonzero inside the metal ball. Question Which one of these statements is false? A. The electric field of a very long uniformly charged rod has a 1/r distance dependence. B. The electric field of a capacitor at a location outside the capacitor is very small compared to the field inside the capacitor. C. The fringe field of a capacitor at a location far away from the capacitor looks like an electric field of a point charge. D. The electric field of a uniformly charged thin ring at the center of the ring is zero. Chapter 17 Electric Potential Potential Energy To understand the dynamics of moving objects we used: forces, momenta, work, energy The concept of electric field E deals with forces Introduce electric potential – for work and energy Electric potential: electric potential energy per unit charge Practical importance: • Reason about energy without having to worry about the details of some particular distribution of charges • Batteries: provide fixed potential difference • Predict possible pattern of E field Energy of a Single Particle The energy of a single particle with charge q1 consists solely of its particle energy. q1 Particle energy Kinetic energy E= mv 2 For v<<c: K » 2 mc 2 v2 1- 2 c = mc 2 + K Rest energy Kinetic energy is associated with motion A single particle has no (electric) potential energy The kinetic energy of a single particle can be changed if positive or negative work is done on the particle by its surroundings. Electric Potential Energy of Two Particles r12 Potential energy is associated with pairs of interacting objects Energy of the system: 1. Energy of particle q1 q2 2. Energy of particle q2 3. Interaction energy Uel q1 Esystem = E1+E2+Uel To change the energy of particles we have to perform work. DE1 + DE2 = Wext + Wint + Q Wext – work done by forces exerted by other objects Wint – work done by electric forces between q1 and q2 Q – thermal transfer of energy into the system Electric Potential Energy of Two Particles D(E1 + E2 ) = Wext + Wint q2 r12 D(E1 + E2 ) - Wint = Wext Uel -Wint q1 if D(mc2 ) = 0 DEsystem = DK system + DUel = Wext + Q Total energy of the system can be changed (only) by external forces or by adding (thermal) energy. Work done by internal forces: f DU el = -Wint = - ò Fint · dr i Electric Potential Energy of Two Particles Fint f DU el = -Wint = - ò Fint · dr q2 r12 i f 1 q1q2 DU el = - ò r̂12 · r̂ dr12 2 4pe 0 r12 i f 1 q1q2 DU el = - ò dr12 2 4pe 0 r12 i q1 1 q1q2 F= rˆ12 2 4pe 0 r12 DU el = - 1 4pe 0 f q1q2 ò i 1 dr12 2 r12 f é 1ù DU el = q1q2 ê- ú 4pe 0 ë r12 ûi 1 Electric Potential Energy of Two Particles Fint r12 q1 é 1ù DU el = q1q2 ê- ú 4pe 0 ë r12 ûi æ 1 q1q2 ö ÷÷ DU el = Dçç è 4pe 0 r12 ø 1 q2 f 1 q1q2 F= rˆ12 2 4pe 0 r12 The potential energy of a pair of particles is: 1 q1q2 U el = (joules) 4pe 0 r12 Electric Potential Energy of Two Particles q2 1 q1q2 U el = (joules) 4pe 0 r12 q1 Uel > 0 for two like-sign charges (repulsion) q2 q1 Uel < 0 for two unlike-sign Charges (attraction) Electric Potential Energy of Two Particles æ 1 q1q2 ö DU el = D ç è 4pe 0 r12 ÷ø Meaning of U0: r12 1 q1q2 U el = + U0 4pe 0 r12 U el ® U 0 Choose U0=0 – no potential energy if r12 (no interaction) q2 q1 Potential energy = amount of work the two charges can do on each other when moved away from each other to q2 q1 Electric and Gravitational Potential Energy q2 1 q1q2 F= r̂ 2 4pe 0 r m1m2 F = -G 2 r̂ r q1 m2 m1 1 q1q2 U el = 4pe 0 r U grav m1m2 = -G r Three Electric Charges Interaction between q1 and q2 is independent of q3 There are three interacting pairs: q1 q2 U12 q2 q3 U23 q3 q1 U31 U= U12+ U23+ U31 1 q1q2 1 q2 q3 1 q1q3 Uel = + + 4pe 0 r12 4pe 0 r23 4pe 0 r13 Multiple Electric Charges q1 q3 q6 q2 q4 Each (i,j) pair interacts: potential energy Uij q5 1 qi q j U el = åU ij = å i< j i < j 4pe 0 rij Notation: i<j avoids double counting: ij, ji Electric Potential Electric potential electric potential energy per unit charge U el V= q Units: J/C = V (Volt) Volts per meter = Newtons per Coulomb Alessandro Volta (1745 - 1827) Electric potential – often called potential Electric potential difference – often called voltage V due to One Particle U el V= q q2 Single charge has no electric potential energy Single charge has potential to interact with other charge – it creates electric potential 1 q1q2 U el = 4pe 0 r 1 q1 VB = 4pe 0 r probe charge J/C, or Volts Electric potential at B due to charge q1. V due to Two Particles Electric potential is scalar: VC = VC ,1 + VC , 2 q1 1 q2 = + 4pe 0 r13 4pe 0 r23 1 Electric potential energy of the system: q3 U sys 1 q1q2 = U12 = 4pe 0 r12 If we add one more charge at position C: U sys 1 q1q2 æ 1 q1 1 q2 ö = U12 + VC q3 = +ç + q3 ÷ 4pe 0 r12 è 4pe 0 r13 4pe 0 r23 ø U sys 1 q1q2 1 q1q3 1 q2 q3 = + + = U12 + U13 + U 23 4pe 0 r12 4pe 0 r13 4pe 0 r23 V at Infinity 1 q1 V= 4pe 0 r r, V=0 Positive charge Negative charge Exercise What is the electrical potential at a location 1Å from a proton? 1Å ( ) -19 2 1.6 ´ 10 C æ ö 1 q1 9 Nm V= = ç 9 ´ 10 = 14.4 J/C = 14.4 V 2 ÷ -10 4pe 0 r è C ø 10 m ( ) What is the potential energy of an electron at a location 1Å from a proton? ( ) U el = Vq= (14.4 J/C) -1.6 ´10-19 C = -2.3´10-18 J = -14.4 eV Exercise 2Å 1Å What is the change in potential in going from 1Å to 2Å from the proton? ∆𝑉 = 𝑉 2Å − 𝑉 1Å = 7.2V − 14.4V = −7.2V What is the change in electric potential energy associated with moving an electron from 1Å to 2Å from the proton? ∆𝑈𝑒𝑙 = 𝑈𝑒𝑙 2Å − 𝑈𝑒𝑙 1Å = 𝑞𝑉 2Å − 𝑞𝑉 1Å = 𝑞Δ𝑉 ( ) DUel = -1.6 ´ 10 -19 C ( -7.2 J/C) = +1.15 ´ 10 -18 J Does the sign make sense? Electric Potential Energy of Two Particles Energy of the system: 1. Energy of particle q1 2. Energy of particle q2 3. Interaction energy Uel q2 r12 To change the energy of particles we have to perform work. DE1 + DE2 = Wext + Wint + Q q1 DE1 + DE2 - Wint = Wext Energy Principle Uel -Wint for a multiparticle system. if D(mc2 ) = 0 : DK system + DUel = Wext + Q Total energy of the system can be changed (only) by external forces. Work done by internal forces: f DU el = -Wint = - ò Fint · dl i Example The electron enters the capacitor through a small hole at A with an initial velocity to the right. What is the change in the electron’s potential energy in traveling from A to B? Change in kinetic energy? (AB)= 4mm; 𝐸 = 2x103N/C. x ∆𝑈𝑒𝑙 = −𝐹𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 ⋅ ∆𝑙 = − −𝑒𝐸𝑥 ∆𝑥 = 𝑒𝐸𝑥 ∆𝑥 = (1.6x10-19 C)(2x103 N/C)(0.004m) =1.3x10-18 J K = -Uelectric = -1.3x10-18 J Electric Potential Difference in a Uniform Field Electric potential electric potential energy per unit charge DU electric DUelectric = qDV, DV = q Units: J/C = V (Volt) Volts per meter = Newtons per Coulomb Electric potential, V – often called potential Electric potential difference, V – often called voltage ∆𝑈𝑒𝑙 = −𝐹𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 ⋅ ∆𝑙 𝑞 𝑞 ∆𝑉 = −𝐸 ⋅ ∆𝑙 Example ∆𝑙 300 ∆𝑉 = −𝐸 ∙ ∆𝑙 = −𝐸∆𝑙 cos(𝜃) 𝑁 = −(100 )(2𝑚) cos 30° 𝐶 = −173 𝑉 V at Infinity 1 q1 V= 4pe 0 r V Positive charge r r, V=0 Electric Potential Energy of Two Particles q2 1 q1q2 U el = (joules) 4pe 0 r12 q1 Uel > 0 for two like-sign charges (repulsion) q2 q1 Uel < 0 for two unlike-sign Charges (attraction) Sign of the Potential Difference DUel = qDV The potential difference V can be positive or negative. The sign determines whether a particular charged particle will gain or lose energy in moving from one place to another. If qV < 0 – then potential energy decreases and K increases If qV > 0 – then potential energy increases and K decreases Path going in the direction of E: Potential is decreasing (V < 0) Path going opposite to E: Potential is increasing (V > 0) Path going perpendicular to E: Potential does not change (V = 0) Sign of the Potential Difference DUel = qDV If freed, a positive charge will move to the area with a lower potential: Vf – Vi < 0 DU el = qDV < 0 DK = -DU el > 0 (no external forces) V1 < V2 Moving in the direction of E means that potential is decreasing Sign of the Potential Difference DUel = qDV To move a positive charge to the area with higher potential: Vf – Vi > 0 DU el = qDV > 0 Need external force to perform work V2 > V1 Moving opposite to E means that potential is increasing Question 1 A proton is free to move from right to left in the diagram shown. There are no other forces acting on the proton. As the proton moves from right to left, its potential energy: A) B) C) D) Is constant during the motion Decreases Increases Not enough information V1 < V2 Potential Difference in a Nonuniform Field C x ∆𝑉 = −𝐸 ∙ ∆𝑙 𝐸1 = 𝐸1𝑥 𝑥 A to C: V1 = -|E1x|(xC-xA) 𝐸2 = − 𝐸2𝑥 𝑥 C to B: V2 = |E2x|(xB-xC); A to B: V = V1+ V2 = -|E1x|(xC-xA) + |E2x|(xB-xC) Potential Difference with Varying Field 𝑥𝑓 𝑓 𝐸 ∙ 𝑑𝑙 = − ∆𝑉 = − 𝑖 𝑦𝑓 𝐸𝑥 𝑑𝑥 − 𝑥𝑖 𝑧𝑓 𝐸𝑦 𝑑𝑦 − 𝑦𝑖 𝐸𝑧 𝑑𝑧 𝑧𝑖 Example: Different Paths near Point Charge 1. Along straight radial path: rf ri f DV = - ò E · dl rf DV = - ò ri i q r̂ · r̂ dr 2 4pe 0 r 1 1 +q q E= r̂ 2 4pe 0 r 1 Origin at +q rf 1 DV = q ò 2 dr 4pe 0 ri r rf é 1ù DV = q ê- ú 4pe 0 ë r û ri 1 é 1 1ù DV = q ê - ú = V f - Vi 4pe 0 êë rf ri úû 1 Example: Different Paths near Point Charge f 2. Special case DV = - ò E · dl i iA: DViA = 0 AB: DVAB é1 1ù = qê - ú 4pe 0 ë r3 r1 û 1 BC: DVBC = 0 𝑞 é1 1ù qê - ú Cf: DVCf = 4pe 0 ë r2 r3 û 1 + DViABCf é1 1 1 1ù 1 é1 1ù = qê - + - ú = q ê - ú = DVif 4pe 0 ë r3 r1 r2 r3 û 4pe 0 ë r2 r1 û 1 Example: Different Paths near Point Charge f DV = - ò E · dl 3. Arbitrary path i E · dl = Edl cosq E · dl = Edlradial f DV = - ò Edlradial i + é 1 1ù DV = q ê - ú = V f - Vi 4pe 0 êë rf ri úû 1