Electric Potential Lecture 5 - PDF Free Download

Chapter 23 Electric Potential Lecture 5 Dr. Armen Kocharian

Electrical Potential Energy When a test charge is placed in an electric field, it experiences a force F = q o E The force is conservative ds is an infinitesimal displacement vector that is oriented tangent to a path through space

Electric Potential Energy, cont The work done by the electric field is F. ds = q o E. ds As this work is done by the field, the potential energy of the charge-field system is changed by ΔU = -q o E. ds For a finite displacement of the charge from A to B, U U U q B Δ = = E ds B A o A

Electric Potential Energy, final Because q o E is conservative, the line integral does not depend on the path taken by the charge This is the change in potential energy of the system

Electric Potential The potential energy per unit charge, U/q o, is the electric potential The potential is independent of the value of q o The potential has a value at every point in an electric field U V = q The electric potential is o

Electric Potential, cont. The potential is a scalar quantity Since energy is a scalar As a charged particle moves in an electric field, it will experience a change in potential ΔU B Δ V = = d q E s A o

Electric Potential, final The difference in potential is the meaningful quantity We often take the value of the potential to be zero at some convenient point in the field Electric potential is a scalar characteristic of an electric field, independent of any charges that may be placed in the field

Work and Electric Potential Assume a charge moves in an electric field without any change in its kinetic energy The work performed on the charge is W = ΔV = q ΔV

Units 1 V = 1 J/C V is a volt It takes one joule of work to move a 1- coulomb charge through a potential difference of 1 volt In addition, 1 N/C = 1 V/m This indicates we can interpret the electric field as a measure of the rate of change with position of the electric potential

Electron-Volts Another unit of energy that is commonly used in atomic and nuclear physics is the electronvolt 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

Potential Difference in a Uniform Field The equations for electric potential can be simplified if the electric field is uniform: B B B A =Δ = E s= s= A A V V V d E d Ed The negative sign indicates that the electric potential at point B is lower than at point A

Energy and the Direction of Electric Field When the electric field is directed downward, point B is at a lower potential than point A When a positive test charge moves from A to B, the charge-field system loses potential energy

More About Directions A system consisting of a positive charge and an electric field loses electric potential energy when the charge moves in the direction of the field An electric field does work on a positive charge when the charge moves in the direction of the electric field The charged particle gains kinetic energy equal to the potential energy lost by the charge-field system Another example of Conservation of Energy

Directions, cont. If q o is negative, then ΔU is positive A system consisting of a negative charge and an electric field gains potential energy when the charge moves in the direction of the field In order for a negative charge to move in the direction of the field, an external agent must do positive work on the charge

Equipotentials Point B is at a lower potential than point A Points A and C are at the same potential The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential

Charged Particle in a Uniform Field, Example A positive charge is released from rest and moves in the direction of the electric field The change in potential is negative The change in potential energy is negative The force and acceleration are in the direction of the field

Potential and Point Charges A positive point charge produces a field directed radially outward The potential difference between points A and B will be 1 1 VB VA keq = r B r A

Potential and Point Charges, cont. The electric potential is independent of the path between points A and B It is customary to choose a reference potential of V = 0 at r A = Then the potential at some point r is V = q ke r

Problem 30. = e q V k r Plot a graph for potential along x direction ( ) V x kq = + = + ( + ) ( + ) e 1 kq e 2 ke Q ke Q r 2 2 2 2 1 r2 x + a x + a ( ) ( ) V x ( ) V x 2kQ e kq e 2 = = 2 2 a 2 x + a ( xa) + 1 = 2 ( kq e a) ( xa) 2 + 1

Problem 30, cont. = e q V k r Now let charge at a to be negative (-Q) and plot along y direction ( ) V y ( + ) ( ) kq = + = + r r y a y+ a e 1 kq e 2 ke Q ke Q 1 2 ( ) V y V ( y) ( kq e a) kq e 1 1 = a ya 1 ya+ 1 1 1 = ya 1 ya+ 1

Electric Potential of a Point Charge The electric potential in the plane around a single point charge is shown The red line shows the 1/r nature of the potential

Electric Potential with Multiple Charges The electric potential due to several point charges is the sum of the potentials due to each individual charge This is another example of the superposition principle The sum is the algebraic sum V = k e i q r V = 0 at r = i i

Electric Potential of a Dipole The graph shows the potential (y-axis) of an electric dipole The steep slope between the charges represents the strong electric field in this region

Today: Electric Potential Energy You should be familiar with the concept of gravitational potential energy from Physics 1 Let's review If a force acts on a particle as the particle moves from a b, then is the work done by the force ( is the infinitesimal displacement along the path)

b a Careful: the force does not necessarily line up with the displacement For example, a block sliding down an inclined plane under the influence of gravity:

Conservative force A force is conservative if the work done by the force is independent of path Only depends on the initial and final points b a Then the work done can be written as function of the difference between some properties of the begin and final point

U is the potential energy W = -ΔU Work energy theorem: work = change in kinetic energy W a b = K(b) K(a) K(a) + U(a) = K(b) + U(b) Potential energy defined up to additive constant

Remember gravitational field, force, potential energy Near the surface of the earth, constant force (F=const) Think of it as mass times constant gravitational field Then gravitational potential energy U=mgh (uniform field) Then gravitational potential energy for two masses is mm 1 2 UG = G r 12

Potential Energy of Multiple Charges Consider two charged particles The potential energy of the system is U = k e qq r 1 2 12

More About U of Multiple Charges If the two charges are the same sign, U is positive and work must be done to bring the charges together If the two charges have opposite signs, U is negative and work is done to keep the charges apart

U with Multiple Charges, final If there are more than two charges, then find U for each pair of charges and add them For three charges: U qq qq qq 1 2 1 3 2 3 = ke + + r12 r13 r23 The result is independent of the order of the charges

Finding E From V Assume, to start, that E has only an x component dv Ex = dx Similar statements would apply to the y and z components Equipotential surfaces must always be perpendicular to the electric field lines passing through them

E and V for an Infinite Sheet of Charge The equipotential lines are the dashed blue lines The electric field lines are the brown lines The equipotential lines are everywhere perpendicular to the field lines

E and V for a Point Charge The equipotential lines are the dashed blue lines The electric field lines are the brown lines The equipotential lines are everywhere perpendicular to the field lines

E and V for a Dipole The equipotential lines are the dashed blue lines The electric field lines are the brown lines The equipotential lines are everywhere perpendicular to the field lines

Electric Field from Potential, General In general, the electric potential is a function of all three dimensions Given V (x, y, z) you can find E x, E y and E z as partial derivatives V V V Ex = Ey = Ez = x y z

Equipotentials and Conductors We argued previously that the electric field must be perpendicular to the surface of a conductor This is because otherwise the charges on the surface of the conductor would be moving It then follows that the surface of a conductor is an equipotential

Potential Gradient We will now derive a fundamental relationship between potential and electric field But This must hold for any (a,b) pair and any path between the two For this to be true then

Write out the components Then, in terms of components: Suppose the displacement is in the x-direction Then dy=dz=0, and dv=e x dx, or

Can do same thing for the other two components Or in short-hand notation is called the "gradient" or the "grad"

If we shift V V + Const. the E-field does not change Makes sense, since V is only defined up-to arbitrary constant The expression above for the gradient is in "Cartesian Coordinates" Cartesian coordinates: x,y,z One important result: If V is a function of r (and not of angle) then

Simple applications of gradient law Point charge: Infinite line of charge Slide 16, this lecture constant

Electric Potential for a 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

V for a Continuous Charge Distribution, cont. To find the total potential, you need to integrate to include the contributions from all the elements dq V = k e r This value for V uses the reference of V = 0 when P is infinitely far away from the charge distributions

V for a Uniformly Charged Ring P is located on the perpendicular central axis of the uniformly charged ring The ring has a radius a and a total charge Q dq dq k = = e V k = e ke dq r 2 2 2 2 x + a x + a

V V for a Uniformly Charged Disk The ring has a radiuses b and a and surface charge density of σ kdq e = = ( 2 2 + ) ( 2 2 x r x + r ) a 1 2rdr V = πk σ = πk σ x + r rdr e b kσ2πrdr e 1 1 2 2 ( 2 2) 2 1 e 2 ( 2 2 + ) 2 x r b ( 2 2) ( 2 2) e 1 1 2 2 V = 2πk eσ x + a 2πk σ x + b a

V V for a Uniformly Charged Disk The ring has a radiuses a and b and surface charge density of σ kdq e = = kσ2πrdr e 1 1 2 2 ( 2 2 + ) ( 2 2 x r x + r ) b 1 2rdr V = πk σ = πk σ x + r rdr e a ( 2 2) 2 1 e 2 ( 2 2 + ) 2 x r a ( 2 2) ( 2 2) e 1 1 2 2 V = 2πk eσ x + b 2πk σ x + a b

V for a Uniformly Charged Disk, cont The ring has a radius a and radius b=0. Surface charge density of σ a ( ) = 2 2 2 V πkeσ x + r 2rdr 0 1 1 ( 2 2) 2 V = 2πk eσ x + a x

V for a Finite Line of Charge A rod of line l has a total charge of Q and a linear charge density of λ l 1 dx Q ( ) = = 2 2 2 V keλ k + = 1 e x r dx 2 2 2 l ( x + a ) 0 0 Q 1 ( 2 2) 2 ke ln x + x + a l 2 2 kq e + + a V = ln a l

V for a Uniformly Charged Sphere A solid sphere of radius R and total charge Q Q For r > R, V = ke r For r < R, E r kq e = 3 R r

V for a Uniformly Charged Sphere For r < R, E = e 3 r r kq e VD VC = Erdr = rdr = 3 R kq e 3 R 2R ( 2 r 2) kq e VD VC = R r 3 2R r R kq R r ( 2 2) kq e kq e kq e VD = + R r = 3R r 3 3 R 2R 2R R V C kq = e R ( 2 2) ( 2 2)

V for a Uniformly Charged Sphere, Graph The curve for V D is for the potential inside the curve It is parabolic It joins smoothly with the curve for V B The curve for V B is for the potential outside the sphere It is a hyperbola

V Due to a Charged Conductor Consider two points on the surface of the charged conductor as shown E is always perpendicular to the displacement ds Therefore, E ds = 0 Therefore, the potential difference between A and B is also zero

V Due to a Charged Conductor, cont. V is constant everywhere on the surface of a charged conductor in equilibrium ΔV = 0 between any two points on the surface The surface of any charged conductor in electrostatic equilibrium is an equipotential surface Because the electric field is zero inside the conductor, we conclude that the electric potential is constant everywhere inside the conductor and equal to the value at the surface

E Compared to V The electric potential is a function of r The electric field is a function of r 2 The effect of a charge on the space surrounding it: The charge sets up a vector electric field which is related to the force The charge sets up a scalar potential which is related to the energy

Irregularly Shaped Objects The charge density is high where the radius of curvature is small And low where the radius of curvature is large The electric field is large near the convex points having small radii of curvature and reaches very high values at sharp points

Irregularly Shaped Objects, cont. The field lines are still perpendicular to the conducting surface at all points The equipotential surfaces are perpendicular to the field lines everywhere

Cavity in a Conductor Assume an irregularly shaped cavity is inside a conductor Assume no charges are inside the cavity The electric field inside the conductor must be zero

Cavity in a Conductor, cont The electric field inside does not depend on the charge distribution on the outside surface of the conductor For all paths between A and B, V B V d A = E s = B A A cavity surrounded by conducting walls is a field-free region as long as no charges are inside the cavity 0

Corona Discharge If the electric field near a conductor is sufficiently strong, electrons resulting from random ionizations of air molecules near the conductor accelerate away from their parent molecules These electrons can ionize additional molecules near the conductor

Corona Discharge, cont. This creates more free electrons The corona discharge is the glow that results from the recombination of these free electrons with the ionized air molecules The ionization and corona discharge are most likely to occur near very sharp points

Millikan Oil-Drop Experiment Experimental Set-Up

Millikan Oil-Drop Experiment Robert Millikan measured e, the magnitude of the elementary charge on the electron He also demonstrated the quantized nature of this charge Oil droplets pass through a small hole and are illuminated by a light

Oil-Drop Experiment, 2 With no electric field between the plates, the gravitational force and the drag force (viscous) act on the electron The drop reaches terminal velocity with F D = mg

Oil-Drop Experiment, 3 When an electric field is set up between the plates The upper plate has a higher potential The drop reaches a new terminal velocity when the electrical force equals the sum of the drag force and gravity

Oil-Drop Experiment, final The drop can be raised and allowed to fall numerous times by turning the electric field on and off After many experiments, Millikan determined: q = ne where n = 1, 2, 3, e = 1.60 x 10-19 C

Van de Graaff Generator Charge is delivered continuously to a high-potential electrode by means of a moving belt of insulating material The high-voltage electrode is a hollow metal dome mounted on an insulated column Large potentials can be developed by repeated trips of the belt Protons accelerated through such large potentials receive enough energy to initiate nuclear reactions

Electrostatic Precipitator An application of electrical discharge in gases is the electrostatic precipitator It removes particulate matter from combustible gases The air to be cleaned enters the duct and moves near the wire As the electrons and negative ions created by the discharge are accelerated toward the outer wall by the electric field, the dirt particles become charged Most of the dirt particles are negatively charged and are drawn to the walls by the electric field

Application Xerographic Copiers The process of xerography is used for making photocopies Uses photoconductive materials A photoconductive material is a poor conductor of electricity in the dark but becomes a good electric conductor when exposed to light

The Xerographic Process

Application Laser Printer The steps for producing a document on a laser printer is similar to the steps in the xerographic process Steps a, c, and d are the same The major difference is the way the image forms on the selenium-coated drum A rotating mirror inside the printer causes the beam of the laser to sweep across the selenium-coated drum The electrical signals form the desired letter in positive charges on the selenium-coated drum Toner is applied and the process continues as in the xerographic process

Potentials Due to Various Charge Distributions

Problem, Method 1. Alternative way to get term is to recognize that there are 4 side pairs and 2 face diagonal pairs. 4 sides of length s and 2 diagonals with length s 2 U = U = U = U U = U 12 23 34 14 13 24 U = 4U + 2U U U 12 13 2 2 e kq e 4kQ 2 1 = + s s 2 2 2 2 kq e kq e = 4+ = 5.41 s 2 s

Problem, Method 2. Alternative way to get term is to recognize that there are 4 side pairs and 2 face diagonal pairs. 4 sides of length s and 2 diagonals with length s 2 U = U + U + U + U 1 2 3 4 ( ) ( ) U = 0+ U + U + U + U + U + U U 12 13 23 14 24 34 2 2 2 e e e kq kq 1 kq 1 = 0+ + 1 1 1 s s + + + + 2 s 2 U 2 2 2 kq e kq e = 4 5.41 s + = 2 s

Problem. Find potential energy. No energy is required for the 20-nC charge placed at its location V 1 ( 9 2 2 8.99 10 N m C ) ( 20 10 9 C ) kq e 1 = = = r 0.04 m ( ) ( ) 9 3 5 U 12 = qv 2 1 = 10 10 C 4.5 10 V = 4.50 10 J ( ) U + U = qv + qv = q V + V 23 13 3 2 3 1 3 2 1. ( ) 5 5 4.50 kv To place the 10-nC charge there we must put in energy Next, to bring up the 20-nC charge requires energy 9 9 9 9 2 2 10 10 C 20 10 C = 20 10 C 8.99 10 N m C + 0.04 m 0.08 m = 4.50 10 J 4.50 10 J 5 U12 + U 23 + U13 = 4.50 10 J

Problem, cont. Find 40 nc charge is released. The three fixed charges create potential at the location where the fourth is released: 9 9 9 9 2 2 20 10 10 10 20 10 V= V1+ V2+ V3= ( 8.99 10 Nm C ) + Cm 2 2 0.04 + 0.03 0.03 0.05 3 V = 3.00 10 V Energy of the system of four charged objects is conserved as the fourth charge flies away: 1 2 1 2 mv + qv = mv + qv 2 2 i ( 1 )( ) ( 2 ) 4 21.20 ( 10 J) 9 3 13 2 0+ 40 10 C 3.00 10 V = 2.00 10 kg v + 0 v = = 3.46 10 m s 13 2 10 kg f 4

Quick Quiz 1 In the figure below, two points A and B are located within a region in which there is an electric field. The potential difference ΔV = V B V A is (a) positive (b) negative (c) zero

Quick Quiz 1 Answer: (b). When moving straight from A to B, E and ds both point toward the right. Thus, the dot product E ds is positive and ΔV is negative.

Quick Quiz 2 In this figure, a negative charge is placed at A and then moved to B. The change in potential energy of the charge field system for this process is (a) positive (b) negative (c) zero

Quick Quiz 2 Answer: (a). From ΔU = q 0 ΔV, if a negative test charge is moved through a negative potential difference, the potential energy is positive. Work must be done to move the charge in the direction opposite to the electric force on it.

Quick Quiz 3 The labeled points of the figure below are on a series of equipotential surfaces associated with an electric field. Rank (from greatest to least) the work done by the electric field on a positively charged particle that moves along the following transitions. (a) A -> B, B -> C, C -> D, D -> E (b) A -> B, D -> E, B -> C, C -> D (c) B -> C, C -> D, A -> B, D -> E (d) D -> E, C -> D, B -> C, A -> B

Quick Quiz 3 Answer: (c). Moving from B to C decreases the electric potential by 2 V, so the electric field performs 2 J of work on each coulomb of positive charge that moves. Moving from C to D decreases the electric potential by 1 V, so 1 J of work is done by the field. It takes no work to move the charge from A to B because the electric potential does not change. Moving from D to E increases the electric potential by 1 V, and thus the field does 1 J of work per unit of positive charge that moves.

Quick Quiz 4 For the equipotential surfaces in this figure, what is the approximate direction of the electric field? (a) Out of the page (b) Into the page (c) Toward the right edge of the page (d) Toward the left edge of the page (e) Toward the top of the page (f) Toward the bottom of the page

Quick Quiz 4 Answer: (f). The electric field points in the direction of decreasing electric potential.

Quick Quiz 5a A spherical balloon contains a positively charged object at its center. As the balloon is inflated to a greater volume while the charged object remains at the center, the electric potential at the surface of the balloon will (a) increase (b) decrease (c) remain the same.

Quick Quiz 5a Answer: (b). The electric potential is inversely proportional to the radius (see Eq. 25.11).

Quick Quiz 5b Recall that the spherical balloon from part a) contains a positively charged object at its center. As the balloon is inflated to a greater volume while the charged object remains at the center, the electric flux through the surface of the balloon will (a) increase (b) decrease (c) remain the same.

Quick Quiz 5b Answer: (c). Because the same number of field lines passes through a closed surface of any shape or size, the electric flux through the surface remains constant.

Quick Quiz 6 In Figure 25.10a, take q 1 to be a negative source charge and q 2 to be the test charge. If q 2 is initially positive and is changed to a charge of the same magnitude but negative, the potential at the position of q 2 due to q 1 (a) increases (b) decreases (c) remains the same

Quick Quiz 6 Answer: (c). The potential is established only by the source charge and is independent of the test charge.

Quick Quiz 7 Consider the situation in question 6 again. When q 2 is changed from positive to negative, the potential energy of the two-charge system (a) increases (b) decreases (c) remains the same

Quick Quiz 7 Answer: (a). The potential energy of the two-charge system is initially negative, due to the products of charges of opposite sign in Equation 25.13. When the sign of q 2 is changed, both charges are negative, and the potential energy of the system is positive.

Quick Quiz 8 In a certain region of space, the electric potential is zero everywhere along the x axis. From this we can conclude that the x component of the electric field in this region is (a) zero (b) in the x direction (c) in the x direction.

Quick Quiz 8 Answer: (a). If the potential is constant (zero in this case), its derivative along this direction is zero.

Quick Quiz 9 In a certain region of space, the electric field is zero. From this we can conclude that the electric potential in this region is (a) zero (b) constant (c) positive (d) negative

Quick Quiz 9 Answer: (b). If the electric field is zero, there is no change in the electric potential and it must be constant. This constant value could be zero but does not have to be zero.

Quick Quiz 10 Consider starting at the center of the left-hand sphere (sphere 1, of radius a) in the figure below and moving to the far right of the diagram, passing through the center of the right-hand sphere (sphere 2, of radius c) along the way. The centers of the spheres are a distance b apart. Draw a graph of the electric potential as a function of position relative to the center of the left-hand sphere.

Quick Quiz 10 Answer: The graph would look like the sketch below. Notice the flat plateaus at each conductor, representing the constant electric potential inside a solid conductor.

Conceptual questions for practice on chapter 23

The positive charge is the end view of a positively charged glass rod. A negatively charged particle moves in a circular arc around the glass rod. Is the work done on the charged particle by the rod s electric field positive, negative or zero? 1. Positive 2. Negative 3. Zero

Rank in order, from largest to smallest, the potential energies U a to U d of these four pairs of charges. Each + symbol represents the same amount of charge. 1. U a = U b > U c = U d 2. U a = U c > U b = U d 3. U b = U d > U a = U c 4. U d > U b = U c > U a 5. U d > U c > U b > U a

A proton is released from rest at point B, where the potential is 0 V. Afterward, the proton 1. moves toward A with an increasing speed. 2. moves toward A with a steady speed. 3. remains at rest at B. 4. moves toward C with a steady speed. 5. moves toward C with an increasing speed.

Rank in order, from largest to smallest, the potentials V a to V e at the points a to e. 1. V a = V b = V c = V d = V e 2. V a = V b > V c > V d = V e 3. V d = V e > V c > V a = V b 4. V b = V c = V e > V a = V d 5. V a = V b = V d = V e > V c

Rank in order, from largest to smallest, the potential differences V 12, V 13, and V 23 between points 1 and 2, points 1 and 3, and points 2 and 3. 1. V 12 > V 13 = V 23 2. V 13 > V 12 > V 23 3. V 13 > V 23 > V 12 4. V 13 = V 23 > V 12 5. V 23 > V 12 > V 13

What are the units of potential difference? 1. Amperes 2. Potentiometers 3. Farads 4. Volts 5. Henrys

New units of the electric field were introduced in this chapter. They are: 1. V/C. 2. N/C. 3. V/m. 4. J/m 2. 5. W/m.

The electric potential inside a capacitor 1. is constant. 2. increases linearly from the negative to the positive plate. 3. decreases linearly from the negative to the positive plate. 4. decreases inversely with distance from the negative plate. 5. decreases inversely with the square of the distance from the negative plate.