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Un importable: #91ADB18D,6 #95A4C6F7,6 #8686DCDC,4 #8280F5F7,6 #8133A715,6 Practice Exam 1 has been posted on course web site 1 hand graded question and 5 multiple choose questions Question 1 (Chap. 16) 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 2 (Chap. 16) 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. Integration Divide shell into rings of charge, each delimited by the angle and the angle + Use polar coordinates (r, ,). R Distance from center: d=(r-Rcos) Rsin Surface area of ring: R Rcos 2 (R sin ) d r 1 (Q)d E 4 0 (d 2 (Rsin )2 )3/2 2 Rsin R Q Q 4 R 2 1 Q (r Rcos ) E sin d 3/2 4 0 2 0 (r Rcos )2 (Rsin )2 A Solid Sphere of Uniformly Distributed Charge What if charges are distributed throughout an object? Step 1: Cut up the charge into shells 1 dQ rˆ For each spherical outside: dE 2 40 r shell: inside: dE = 0 Outside a solid sphere of charge: E 1 Q rˆ 2 40 r for r>R R r E A Solid Sphere of Uniformly Distributed Charge Inside a solid sphere of charge: E 1 Q rˆ 2 40 r (volume of inner shells) Q Q (volume of sphere) 4 / 3 r 3 r3 Q Q Q 3 3 4 / 3 R R E 1 Qr r̂ 3 4 0 R for r<R Why is E~r? On surface: 1 QR 1 Q E r̂ r̂ 3 2 4 0 R 4 0 R R r E Chapter 17 Electric Potential 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? ( ) -19 -18 = 14.4 J/C -1.6 ´ 10 C = -2.3 ´ 10 J ( ) U el = Vq Exercise 2Å 1Å What is the change in potential in going from 1Å to 2Å from the proton? DV = V æ 2 Aö - V æ 1Aö = -7.2 V è ø è ø What is the change in electric potential energy associated with moving an electron from 1Å to 2Å from the proton? DUel = U el æ 2 Aö - U el æ 1Aö = qV æ 2 Aö - qV æ 1Aö = qDV è ø è ø è ø è ø ( ) DUel = -1.6 ´ 10 -19 C ( -7.2 J/C) = +1.15 ´ 10 -18 J Does the sign make sense? Potential Energy To understand the dynamics of moving objects consider: forces, momenta, work, energy Introduced the concept of electric field E to deal with forces Introduce electric potential – to consider 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 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. Work done by internal forces: f DU el = -Wint = - ò Fint · dr i 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 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 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 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 Question 1 What is the potential in the center of uniformly charged hollow sphere? ([k=1/(40)]) Q R A) B) C) D) E) k*Q/R2 k*Q/R k*Q/(4R2) 0 Not enough information Potential Inside a Uniformly Charged Hollow Sphere A A V A E dl E dl E dr A 1 Q VA = 4pe 0 R =0 B A VB E dl ( E dl E dl ) B 1 Q VB = 4pe 0 R A