Electrostatic Potential

Chapter 23 Electrostatic Potential

PowerPoint® Lectures for University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman

Lectures by James Pazun Modified by P. Lam 7_11_2008

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Topics for Chapter 23 • Electrostatic potential energy • Electrostatic potential function, V(r), generated by source charges • Connection between electrostatic potential, V(r), and the electrostatic field, E(r), generated by source charges • Intermission • Find V(r) and E(r) for continuous charge distributions • Equipotential surfaces

• When two masses are separated (such as a baseball and the Earth) by a distance r, we say that there is a gravitational potential energy associated with this configuration because if we let go the masses, they will move (toward each other); potential energy is converted into kinetic energy. • Similarly, if two charges are separated by a distance r, there is a electrostatic potential energy associated with this configuration, if we let go the charges, they will move (away from each other for like charges; toward each other for opposite charges); potential energy is again converted into kinetic energy.

m 2 q q 2 + 2 r + r r OR

m2 q q1 + 1 -

m q2 q2 2 + + r r r OR

m2 q q1 + 1 - Electrostatic potential energy: kq q Gravitational potential energy: U = 1 2 (q's can be positive or negative) e r Gm m U = 1 2 For two unlike charges, g r lower potential energy if charges are closer Lower potential energy (they like to move closer) if the masses are closer. For two like charges, (potential energy is more negative) Masses tends to move toward lower potential energy if lower potential energy if they charges are further apart are free to move. (they like to move away) Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Relating work done by electrostatic force to change in electrostatic potential energy q q2 2 Example : Given : q1 =+2C,q2 =+1C,ri = 3m + + m =10kg,m = 5kg. r 1 2 r Let go of q2 and it moves to infinity.

OR Find the final kinetic energy of q2.

Find the final speed of q2. q q1 + 1 - f r r f kq q Challengequestion : Let go both q1 and q2; W= F dl 1 2 dr • = 2 they both move to infinity. Find the final speeds i i r of q1 and q2. kq1q2 kq1q2 = = ()U f Ui rf ri Work Kinetic Energy Theorem :

K f Ki = W = ()U f Ui

K f + U f = Ki + Ui (Conservation of energy) (Electrostatic force is a conservative force r dU F = rˆ ) dr Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Electrostatic potential function

q1 q2

r Electric field concept : r r r q1 generates E (r ) everywhere q2 at location r r r r r r senses E (r ) and experiences a force F = q2E( r )

Electrostatic potential (V(r )) concept : r r q1 generates V(r ) everywhere q2 at location r senses V(r ) and the potential energy of this interaction r is U = q2V(r )

r kq1 kq1 If q1 is a point charge, then V(r ) = = ; r x 2 + y 2 + z2 the origin is chosen to be the location of q . 1 Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Connection between electrostatic potential and electrostatic field

q1 q2 Use point charges as example r r r dU Note :V (r ) is a scalar (not a vector) r F = rˆ so, it is easier to calculate than E (r ). dr r r r Once V(r ) is known, E(r ) can be r F dU()/q dV calculated from V(r ). E = = 2 rˆ = rˆ q2 dr dr kq r dV kq check :V (r ) = 1 E rˆ = 1 rˆ r dr r2 kq If we express V(r ) = V(x,y,z) = 1 x 2 + y 2 + z2 r V V V ˆ then E = i ˆ + ˆj + k x y z

Potential energy of source charges q1,q2,andq3

kq1q2 kq1q3 kq2q3 Usource = + + (i.e. all possible pair combinations) ch arg es r12 r13 r23

If a test charge qo is added, then the additional potential energy is kq1 kq2 kq3 Usourcetest = q0 + + qoVsource charge interaction r1 r2 r3 That is, the electrostatic potential generated by several source charges is the sum of the individual potentials.

Given : q1 =+12nC,q2 = 12nC (1)Find the electrostatic potential at points a, b , and c (2) Is the potential at point c zero? Is the electric field at point c zero?

(3) A test charge qo =+1nC travels from point b to point c, how much work is done by the electrostatic force?

(1) Calculate V(x,y = 0,z = 0) for all x. (2) From the potential function V(x,y = 0,z = 0),

find E x (x,y = 0,z = 0) r (3) How would you find E at any point?

y (1) Calculate V(x,y) for all x and y kdQ V(x,y,z) = ; dQ = dy' dQ r dy’ 2 2 2 r = x + (y y') + z a kdy' V(x,y,z) = 2 2 2 a x + (y y') + z P=(x,y,z) y’ Once this is calculated, then we can r x find E(x,y,z) by taking derivatives, r V V V ˆ E (x,y,z)= i ˆ + ˆj + k x y x You are NOT required to do the integrals, but need to understand the concept.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Calculation of electrical potential from E-field • If the E-field is known (say using Gauss’s Law), then one find the potential from E(r).

r dV r r E(r) = - rˆ V(r) = E (r')• dr ' dr

A positive test charge (if let go) would move from a region of higher potential to V=7volts a region of lower potential (fundamentally it moves from high potential V=3volts energy to low potential energy; U = qV)

A negative test charge (if let go) would move from a region of lower potential to a region of higher potential (again, fundamentally it moves from high potential V=7volts energy to low potential energy; U = - | q | V)

V=3volts

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Equipotential surfaces and field lines • Surfaces of equal potential are perpendicular to the field lines. • Spacing between equi-potentials => strength of E-field • closer spacing => higher E-field strength

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Field lines and a conducting surface • Refer to Figure 23.25 to illustrate the concept of field lines near a conducting surface.