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Physics 2113 Jonathan Dowling. Physics 2113 Lecture: 14. Electric Potential III. Work Done (W) is Integral of Force (F). Potential Energy (U) is Negative of Work Done. Hence Force is Negative Derivative of Potential Energy. Conservative Forces, Work, and Potential Energy. Electric
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Physics 2113 Jonathan Dowling Physics 2113 Lecture: 14 Electric Potential III
Work Done (W) is Integral of Force (F) Potential Energy (U) is Negative of Work Done Hence Force is Negative Derivative of Potential Energy Conservative Forces, Work, and Potential Energy
Electric Field [N/C]=[V/m] Force [N] = Newton Potential Energy [J]=Joule Potential Voltage [J/C]=[V] =Volt Coulomb’s Law for Point Charge q2 q2 q1 P1 P2 P1 P2
Potential At Center of Ring of Charge r • Divide the charge distribution into differential elements • Write down an expression for potential voltalge from a typical element — treat as point charge • Integrate! • Simple example: circular rod of radius r, total charge Q; find V at center. dq Same result holds along z-axis!
Potential Along Axis of Ring of Charge R’ • Divide the charge distribution into differential elements • Write down an expression for potential voltalge from a typical element — treat as point charge • Integrate! dq dq
dq=λdx dq a L Potential Voltage of Continuous Charge Distribution: Return of the Rod! • Uniformly charged rod • Total charge +Q • Length L • What is V at position P? r = x P Units: [Nm2/C2][C/m]=[Nm/C]=[J/C]=[V]
dq=λdx a L Potential Voltage of Continuous Charge Distribution: Return of the Rod! • Uniformly charged rod • Total charge +Q • Length L • What is V at position P? • What about a>>L? r = x P In this limit we recover V=kq/r for point charge!
Potential Voltage of Continuous Charge • Uniformly charged rod • Total charge +Q • Length L • What is V at position P?
Potential Voltage of Continuous Charge • What about d>>L? We recover formula V=kq/r for a point charge Again!
Electric Field & Potential: A Simple Relationship! Focus only on a simple case: electric field that points along +x axis but whose magnitude varies with x. Notice the following: • Point charge: E = kQ/r2 V = kQ/r • Dipole (far away): E = kp/r3 V = kp/r2 • E is given by a DERIVATIVE of V! • Of course! • Note: • MINUS sign! • Units for E: VOLTS/METER (V/m)
dq a L Units: √ Electric Field! E from V: Example • Uniformly charged rod • Total charge +Q • Length L • We Found V at P! • Find E from V? x P
Electric Field & Potential: ICPP (b) V V (a) r r r=R r=R R • Hollow metal spherical shell of radius R has a charge on the shell +q • Which of the following is magnitude of the electric potential voltage V as a function of distance r from center of sphere? Hint: Inside sphere there are no charges so E=0. But E=dV/dr=0. What can V be?
(a) Since Δx is the same, only |ΔV| matters! |ΔV1| =200, |ΔV2| =220, |ΔV3| =200 |E2| > |E3| = |E1| The bigger the voltage drop the stronger the field. Δx (b) = 3 (c) F = qE = ma accelerate leftward
Equipotentials and Conductors V E • Conducting surfaces are EQUIPOTENTIALs • At surface of conductor, E is normal (perpendicular) to surface • Hence, no work needed to move a charge from one point on a conductor surface to another • Equipotentials are normal to E, so they follow the shape of the conductor near the surface.
An Uncharged Conductor: A Uniform Electric Field: An Uncharged Conductor in the Initially Uniform Electric Field: Conductors Change the Field Around Them!
Sharp Conductors • Charge density is higher at conductor surfaces that have small radius of curvature • E =σ/ε0for a conductor, hence STRONGER electric fields at sharply curved surfaces! • Used for attracting or getting rid of charge: • lightning rods • Van de Graaf -- metal brush transfers charge from rubber belt • Mars pathfinder mission -- tungsten points used to get rid of accumulated charge on rover (electric breakdown on Mars occurs at ~100 V/m) (NASA)
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Summary: • Electric potential: work needed to bring +1C from infinity; units = V = Volt • Electric potential uniquely defined for every point in space -- independent of path! • Electric potential is a scalar -- add contributions from individual point charges • We calculated the electric potential produced by a single charge: V = kq/r, and by continuous charge distributions : dV = kdq/r • Electric field and electric potential: E= -dV/dx • Electric potential energy: work used to build the system, charge by charge. Use W = U = qV for each charge. • Conductors: the charges move to make their surface equipotentials. • Charge density and electric field are higher on sharp points of conductors.