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Chapter 29 Electric Potential Reading: Chapter 29 1 Electrical Potential Energy When a test charge is placed in an electric field, it experiences a force E F qE q s If s is an infinitesimal displacement of test charge, then the work done by electric force during the motion of the charge is given by sF qsE 2 Electrical Potential Energy sF qsE E F qE q This is the work done by electric field. s In this case work is positive. g F mg s This is very similar to gravitational force: the work done by force is Because the positive work is done, the potential energy of charge-field system should decrease. So the change of potential energy is U sF qsE sF mgs The change of potential energy is U sF mgs sign minus 3 Electrical Potential Energy: Example F qE A q α s B UB UA sF qsE qsE cos α E UB UA qsE cos α A q F qE α s UB UA s1F s2F qs1E s1 s2 B s2E 0 s1 s cos α UB UA qsE cos α E The change of potential energy does not depend on the path: The electric force is conservative 4 Electrical Potential Energy F qE α q A ds s E ds B B U UB UA q E ds A ds is oriented tangent to a path through space For all paths: UB UA qsE cos α The electric force is conservative 5 Electric Potential Electric potential is the potential energy per unit charge, V U q The potential is independent of the value of q. The potential has a value at every point in an electric field Only the difference in potential is the meaningful quantity. UB U A sF VB VA sE sE cos α q q q A q F qE α s B E 6 Electric Potential To find the potential at every point 1. we assume that the potential is equal to 0 at some point, for example at point A, VA 0 2. we find the potential at any point B from the expression B B A A VB VA Eds Eds A VB sE cos α α s B E 7 Electric Potential: Example Plane: Uniform electric field σ 0 VA 0 σ E ε B B E h ds B σh VB VA Eds E ds hE 2ε 0 A A A σ 0 E ds h C C C σh VC VA Eds E ds hE 2ε 0 A A 8 Electric Potential: Example V σ 0 h V σ Plane: Uniform electric field B E h σh VB 2ε 0 h 0 A σ E h C σh VC 2ε 0 All points with the same h have the same potential 9 Electric Potential: Example Plane: Uniform electric field The same potential σh V 2ε 0 equipotential lines σh V 2ε 0 10 Electric Potential: Example Point Charge V 0 V 0 Q Er ke 2 r ds B E r Q0 B r r r dr 1 Q VB VA Eds Er dr keQ 2 keQ ke r r r 11 Electric Potential: Example Point Charge V 0 equipotential lines Q Vr ke r 12 Electric Potential: Example Point Charge Q Vr ke r V 0 • The potential difference between points A and B will be B 1 1 VB VA keQ Eds rB rA A Q 13 Units • Units of potential: 1 V = 1 J/C – V is a volt – It takes one joule (J) of work to move a 1-coulomb (C) charge through a potential difference of 1 volt (V) Another unit of energy that is commonly used in atomic and nuclear physics is the electron-volt One electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 volt 1 eV = 1.60 x 10-19 J 14 Potential and Potential Energy • If we know potential then the potential energy of point charge q is U qV (this is similar to the relation between electric force and electric field) F qE 15 Potential Energy: Example What is the potential energy of point charge q in the field of uniformly charged plane? U QV Q σh V 2ε 0 h σ σQ 0 U 0 σQ U qV h 2ε 0 h σQ 0 U 0 repulsion h attraction 16 Potential Energy: Example What is the potential energy of two point charges q and Q? This can be calculated by two methods: Uq qVQ A q Q r Q VQ ke r UQ QVq B q Q r q Vq ke r The potential energy of point charge q in the field of point charge Q qQ Uq qVQ ke r The potential energy of point charge Q in the field of point charge q qQ UQ QVq ke r In both cases we have the same expression for the energy. This expression gives us the energy of two point charges. qQ U ke r 17 Potential Energy: Example Potential energy of two point charges: q Q U ke r U qQ r U qQ 0 repulsion qQ 0 attraction 0 r 0 r 18 Potential Energy: Example Find potential energy of three point charges: q2 q1 r12 r13 r23 U12 ke U U12 U13 U23 q1q2 r12 U13 ke q1q3 r13 U23 ke q2q3 r23 q3 U U12 U13 U23 ke qq qq q1q2 ke 1 3 ke 2 3 r12 r13 r23 19 Potential Energy: Applications: Energy Conservation For a closed system: Energy Conservation: The sum of potential energy and kinetic energy is constant q1 q2 r12,initial r23,initial r13 U U12 U13 U23 - Potential energy mv 2 T 2 - Kinetic energy r12,final q3 r23,final q2 Example: Particle 2 is released from the rest. Find the speed of the particle when it will reach point P. v2 Initial Energy is the sum of kinetic energy and potential energy (velocity is zero – kinetic energy is zero) Einitial T U T U12 U13 U23 ke q1q2 r12,initial q1q3 q2q3 ke ke 20 r13 r23,initial Potential Energy: Applications: Energy Conservation For a closed system: Energy Conservation: The sum of potential energy and kinetic energy is constant q1 q2 r12,initial r23,initial r13 r12,final q3 r23,final q2 v2 Final Energy is the sum of kinetic energy and potential energy (velocity of particle 2 is nonzero – kinetic energy) Efinal T U T U12 U13 U23 qq qq m2v 22 qq ke 1 2 ke 1 3 ke 2 3 2 r12,final r13 r23,21final Potential Energy: Applications: Energy Conservation For a closed system: Energy Conservation: q1 The sum of potential energy and kinetic energy is constant q2 r12,initial r23,initial r13 r12,final q2 Final Energy = Initial Energy q3 ke q1q2 r12,initial r23,final v2 q1q3 q2q3 q1q3 q 2 q3 m2v 22 q1q2 ke ke ke ke ke r13 r23,initial 2 r12,final r13 r23,final q2q3 q2q3 m2v 22 q1q2 q1q2 ke ke ke ke 2 r12,initial r12,final r23,initial r23,final v2 2 m2 1 1 k q q e 1 2 r12,initial r12,final 1 1 k q q e 2 3 r23,initial r23,final 22 Electric Potential: Continuous Charge Distribution 23 Electric Potential of Multiple Point Charge Q Vr ke r P q4 q1 q2 q3 q3 q1 q2 q4 Vr V1 V2 V3 V4 ke ke ke ke r1 r2 r3 r4 The potential is a scalar sum. The electric field is a vector sum. 24 Electric Potential of Continuous Charge Distribution • Consider a small charge element dq – Treat it as a point charge • The potential at some point due to this charge element is dq dV ke r • To find the total potential, you need to integrate to include the contributions from all the elements V 0 dq V ke r The potential is a scalar sum. 25 The electric field is a vector sum. Spherically Symmetric Charge Distribution Uniformly distributed charge Q Q V ? 26 Spherically Symmetric Charge Distribution Two approaches: “Complicated” Approach A: dq V ke r “Simple” Approach B: B VB VA Eds (simple - only because we know E(r)) 27 Spherically Symmetric Charge Distribution Q Er ke 3 r a r a Q Er ke 2 r r a 28 Spherically Symmetric Charge Distribution V 0 r a V 0 ds Q0 B E Q Er ke 2 r r B r r r dr 1 Q VB VA Eds Er dr keQ 2 keQ ke r r r 29 Spherically Symmetric Charge Distribution V 0 V 0 r a Er ke ds Q r a3 r Q Er ke 2 r E C B Q0 B a r r a VB VA Eds Er dr Er dr Er dr a 2 Q dr Q Q Q r 2 2 ke 3 rdr keQ 2 ke 3 (a r ) ke ke 3 2 a r r 2a a 2a a 30 r Spherically Symmetric Charge Distribution Vr ke Q r Q r2 VB ke 3 2 2a a r a r a 31 Important Example E 0 σ E ε E 0 σ 0 E σ 0 σ 0 V ? σ 0 32 Important Example A VA 0 ds B B E 0 σ E ε E 0 VB Eds 0 A σ 0 E σ 0 33 Important Example A VA 0 ds E 0 σ 0 C σ E ε x h E B σ 0 E 0 B C B B A A C C VB Eds Eds Eds Eds Ex 34 Important Example A VA 0 ds E 0 σ 0 C σ E ε E h D σ 0 E 0 B B C D B D A A C D C VB Eds Eds Eds Eds Eds Eh 35 Important Example V 0 E 0 σ E E ε σ 0 x σ 0 E 0 σ V x ε σ V Eh h ε V h x σ h ε 36 q2 q1 r12 r13 U U12 U13 U23 r23 q1q3 q2q3 q1q2 ke ke ke r12 r13 r23 q3 r a Q Vr ke r r a Q r2 VB ke 3 2 2a a 37