Physics 121 - Electricity and Magnetism Lecture 05 -Electric Potential Y&F Chapter 23 Sect. 1-5 Electric Potential Energy versus Electric Potential Calculating the Potential from the Field Potential due to a Point Charge Equipotential Surfaces Calculating the Field from the Potential Potentials on, within, and near Conductors Potential due to a Group of Point Charges Potential due to a Continuous Charge Distribution Summary

Electrostatics: Two metal spheres, different radii, one charged Initially Connect wire between spheres, then disconnect it wire Q10= 10 C r1= 10 cm Q1f = ?? Q2f = ?? Are final charges equal? What determines how charge redistributes itself? Q20= 0 C r2= 20 cm X P1 = rgy1 P2 = rgy2 Open valve, water flows. What determines final water levels? Mechanical analogy: Water pressure PE/unit mass = gy Pressure = rgy Potential energy of water molecule at the top surface = mgy

ELECTROSTATIC POTENTIAL Define: Electrostatic Potential equals Potential Energy per unit (test) charge Potential Energy…… DU = Work done ~ force x displacement or charge x field x displacement. Potential…. DV = Work done on a test charge/unit charge ~ field x displacement. = Change in PE of a test charge as it is displaced / test charge magnitude Units, Dimensions: potential Energy DU: Joules potential DV: [U]/[q] Joules/C. = VOLTS units of E field are [V]/[d] = Volts / meter – same as N/C. Potential: is a scalar field  easier to use than E (vector field). E = Gradient (Potential). summarizes effect of distant charges without specifying a test charge there. Both DU and DV: use a reference level (possibly at infinite distance) represent conservative forces (scalar fields) like gravity. can determine motion of charged particles using: Second Law, F = qE or PE, Work-KE theorem &/or mechanical energy conservation.

Reminder: Work Done by a Constant Force 5-1: In the four examples shown in the sketch, a force F acts on an object and does work. The force has the same magnitude in each case. The displacement of the object is to the right and also has the same magnitude. Rank the cases in order of the work done by the force on the object, from most positive to the most negative. I II III IV Ds I, IV, III, II II, I, IV, III III, II, IV, I I, IV, II, III III, IV, I, II Force may vary in direction and magnitude along the path: Result of a “Path Integration” in general can depend on integration path taken

(from basic definition) Definitions: Electrostatic Potential versus Potential Energy E field is “conservative”, like gravity Work done BY THE FIELD on a test charge moving from i to f does not depend on the path taken. Work done around any closed path equals zero. Work done changes potential energy and potential. POTENTIAL ENERGY DIFFERENCE: Charge q0 moves from i to f along ANY path path integral POTENTIAL DIFFERENCE: Potential is work done by field per unit charge ( Evaluate integral on ANY path from i to f ) (from basic definition)

Summary and details Focus on differences in electric potential and PE. Reference levels can be: Relative: Choose arbitrary zero reference level for ΔU or ΔV. …or… Absolute: Set Ui = 0 for all charges that started infinitely far apart Volt (V) = SI Unit of electric potential 1 volt = 1 joule per coulomb = 1 J/C 1 J = 1 VC and 1 J = 1 N m Electric field units – new name: 1 N/C = (1 N/C)(1 VC/1 Nm) = 1 V/m “Electron volt” is an energy unit: 1 eV = work done moving charge e through a 1 volt potential difference = (1.60×10-19 C)(1 J/C) = 1.60×10-19 J The field is created by a charge distribution somewhere else. A test charge q0 moved between i and f gains or loses potential energy DU. DU and DV do not depend on path. DV also does not depend on |q0| (test charge). Use Work-KE theorem to link potential differences to motion.

Work and PE : Who/what does positive or negative work? 5-2: In the figure, we exert an external force that moves a proton from point i to point f in a uniform electric field directed as shown. Which of the following statements is true? E i f A. Electric field does positive work on the proton. Electric potential energy of the proton increases. B. Electric field does negative work on the proton. Electric potential energy of the proton decreases. C. Our external force does positive work on the proton. D. Our external force does positive work on the proton. E. The changes cannot be determined. Hint: which directions pertain to displacement and force?

DV or DU depend only on the endpoints EXAMPLE: Find change in potential DV as test charge +q0 moves from point i to f in a uniform field E i f o Dx DV or DU depend only on the endpoints ANY PATH from i to f gives same results To convert potential to PE just multiply by q0 EXAMPLE: CHOOSE A SIMPLE PATH THROUGH POINT “O” Displacement i  o is normal to field (path along equipotential) External agent must do positive work on positive test charge to move it from o  f E field does negative work Units of E are also volts/meter What are signs of DU and DV if test charge is negative?

Find potential V(R) a distance R from a point charge q : Potential Function for a Point Charge Reference level: V(r) = 0 for r = infinity Field is conservative  choose radial integration path for E(r) DU = q0DV = work done on a test charge as it moves to final location Find potential V(R) a distance R from a point charge q : Positive for q > 0, Negative for q < 0 Inversely proportional to r1 NOT r2 For potential ENERGY: use same method but integrate force Shared PE between q and Q Overall sign depends on both signs

Visualizing the potential function V(r) for a positive point charge (2 D) For q negative V is negative (funnel) r V(r) 1/r

- Electric field lines are perpendicular to the equipotentials On an “Equipotential Surface” (Contour) - Voltage and potential energy are constant; i.e. DV=0, DU=0 q Vfi Vi > Vf DV = 0 (E is the gradient of V) DV = 0 if Ds is normal to E - Zero work is done moving charges along an equipotential and - Electric field lines are perpendicular to the equipotentials

so DV = 0 along any path on or in a conductor CONDUCTORS ARE ALWAYS EQUIPOTENTIALS - Charge on conductors moves to make Einside = 0 - Esurf is perpendicular to surface - Charge is on the outer surface so DV = 0 along any path on or in a conductor Demonstration Source: Pearson Study Area - VTD Chapter 23 Charged Conductor with Teardrop Shape Where on the surface is the field most strong? https://mediaplayer.pearsoncmg.com/assets/secs-vtd34_chargeovoid Discussion: Blunt and sharp ends are at the same potential How do the potential, field, and surface charge densities differ for the blunt and sharp ends? E = s / e0 just outside the surface and is outward At the sharp end s is larger, so E is larger

Distinction: Slope, Grade, Gradients in Gravitational Field EXTRA TOPIC Height contours portray constant gravitational potential energy DU = mgDh. Force is along the gradient, perpendicular to a potential energy contour. Grade means the same thing as slope. A 15% grade is a slope of 100 15 15% q Dh Dl Dx Gradient is measured along the path. For the case at left it would be: The gradient of the potential energy is the gravitational force component along path Dl: The GRADIENT of height (or gravitational potential energy) is a vector field representing steepness (or gravitational force) In General: The GRADIENTS of scalar fields are vector fields.

The field E(r) is the gradient of the potential EXTRA TOPIC q equipotentials E Component of ds on E changes potential Component of ds normal to E produces no change Field is normal to equipotential surfaces For a path along equipotential, DV = 0 E = -spatial rate of change of V = -Gradient of V

Examples of Equipotential Surfaces Point charge or outside a sphere of charge Uniform Field Dipole Field Equipotentials are planes (evenly spaced) Equipotentials are spheres (spaced for 1/r) Equipotentials are not simple shapes

Potential difference between oppositely charged conductors (parallel plate capacitor) Dx E = 0 Vf Vi Assume infinite sheets (Dx << L) Equal and opposite surface charges All charge moves to inner surfaces (opposite charges attract) A positive test charge +q gains potential energy DU = qDV as it moves from - plate to + plate along any path (including an external circuit) Example: Find the potential difference DV across the capacitor, assuming: s = 1 nanoCoulomb/m2 Dx = 1 cm & points from negative to positive plate Uniform field E (from + to – plate)

Conductors are always equipotentials Example: Two spheres, different radii, one charged to 90,000 V. Connect wire between spheres – charge moves Conducting spheres come to same potential Net charge on system is constant, but redistributes to make it an equipotential. Spherical symmetry. r1= 10 cm r2= 20 cm V10= 90,000 V. V20= 0 V. Q20= 0 V. wire Initially (use shell theorem): Find the final charges: Find the final potential(s):

Potential inside a hollow conducting shell Find potential Va at point “a” inside shell R a c b d Assume Vb = 18,000 Volts on surface. Vc = Vb (conducting shell is an equipotential). Shell can be any closed surface (sphere or not) Definition: Apply Gauss’ Law with GS just inside shell: E(r) r Einside=0 R Eoutside= kq / r2 V(r) Vinside=Vsurf Voutside= kq / r Potential is continuous across conductor surface – field is not

Potential due to a group of point charges Use superposition for n point charges Potential is a scalar… so … the sum is an algebraic sum, not a vector sum. Comparison: For the electric field… E may be zero where V does not equal to zero. V may be zero where E does not equal to zero.

Examples: potential due to point charges Use Superposition Note: E may be zero where V does not = 0 V may be zero where E does not = 0 TWO EQUAL CHARGES – Point P at the midpoint between them +q P d F and E are zero at P but work would have to be done to move a test charge to P from infinity. DIPOLE – Otherwise positioned as above +q -q P d

Find E & V at center point P Examples: potential due to point charges - continued Another example: square with charges on corners q -q a d Find E & V at center point P P Another example: same as above with all charges positive

Electric Field and Electric Potential 5-3: Which of the following figures have both V=0 and E=0 at the red point? A q q -q B q C D q -q E -q q

Rate of potential change perpendicular to equipotential Method for finding potential function V of a continuous charge distribution, at point “P” 1. Assume V = 0 infinitely far away from charge distribution (finite size) 2. Find an expression for dq, the charge in a “small” chunk of the distribution, in terms of l, s, or r Typical challenge: express above in terms of chosen coordinates 3. At point P, dV is the differential contribution to the potential due to a point-like charge dq located in the distribution. Use symmetry to simplify. 4. Use “superposition”. Add up (integrate) the contributions over the whole charge distribution, varying the displacement r as needed. Scalar VP. 5. Field E can be gotten from potential by taking the “gradient”: Rate of potential change perpendicular to equipotential

Example 23.11: Potential along Z-axis of a ring of charge y z r P a f dq Q = charge on the ring l = uniform linear charge density = Q/2pa r = [a2 + z2]1/2 = distance from dq to “P” ds = arc length = adf All scalars - no need to worry about direction As z  0, V  kQ/a As a  0 or z  inf, V  point charge Looks like a point charge formula, but r is on the ring FIND ELECTRIC FIELD USING GRADIENT (along z by symmetry) E  0 as z  0 (for “a” finite) E  point charge formula for z >> a As Before

Example 23.10: Potential near an infinitely long charged line (or a charged conducting cylinder) Near line or outside cylinder at r > R E is radial. Choose radial integration path if Above is negative for rf > ri with l positive Inside conducting cylinder for r < R E is radial. Choose radial integration path if Potential inside is constant and equals surface value

Example 23.12: Potential at symmetry point P near a finite line of charge Uniform linear charge density Charge in length dy Potential of point charge From standard integral tables: Limiting cases: Point charge formula for x >> 2a Example 23.10 formula for near field limit x << 2a Use infinite series approximation

Optional Example: Potential Due to a Charged Rod A rod of length L located parallel to the x axis has a uniform linear charge density λ. Find the electric potential at a point P located on the y axis a distance d from the origin. Start with The following will be used below to integrate: Now, let me give you an example to show how to do that. Let’s assume the rod is lying along the x axis, dx is the length of one small segment, and dq is the charge on that segment. So we have dq=lamda times dx. The magnitude of the electric field at P due to this small segment is The total field at P is the integration from a to l+a. Since lamda equals to Q/l, the electric field can be expressed like that. Now, let’s see two extreme cases: (1) l=>0, means the rod has shrunk to zero size,, equation reduces to the electric field due to a point charge. (2) a>>l, means P is far from the rod, then l in the denominator can be neglected. This is exactly the form we would expect for a point charge. Integrate over the charge distribution using interval 0 < x < L Result

Homework Example: Potential on the symmetry axis of a charged disk z q P a dA=adfda r Q = charge on disk whose radius = R. Uniform surface charge density s = Q/pR2 Disc is a set of rings, radius a, each of them da wide in radius For one of the rings: Double integral Integrate twice: first on azimuthal angle f from 0 to 2p which yields a factor of 2p then on ring radius a from 0 to R Use Anti-derivative:

Limiting cases: Potential on the symmetry axis of a charged disk z q P a dA=adfda r An approximation using infinite series “Far field”: assume z>>R Disc mimics a point charge far away “Near field”: assume z<<R Disk mimics an infinite non-conducting sheet of charge