Thursday, February 11, 1999
Announcements: none

Lecture notes:

• Let us consider the case of a test charge, q_o, in an electrostatic field E.
• F = q_oE

• Work done by the electrostatic force on the test charge for a displacement, ds, is dW = F (dot product) ds = q_oE (dot product) ds = - dU
• dU = - q_oE (dot product) ds
• DU = U_f - U_i = -q_o I( Integral from i to f) E (dot product) ds
• i and f mean initial and final position
• The above expression for the change in potential energy is true also for the case when the electric field is not constant; the integral can be evaluated of we know the relation of  E to s.
• Since q_oE  is a conservative force, DW, DU is independent of the path taken from point i to f

• If the charge q_o is brought through a closed loop ( from i to f back to i), then the change in potential energy is equal to zero; the net work done is also zero.

• The electric potential difference, V_f - V _i = ( U_f - U _i )/ q_o= - I( Integral from i to f) E (dot product) ds
• 1 volt = 1 joule per coulomb
• Note that q_oE = F, and the units of E is Newton/ Coulomb or N/C
• Therefore E can be expressed in units of volt/ meter
• 1 N/C = 1 volt/m
• DV = DU / q_o, therefore  DU = DV q_o
• 1 eV is the energy gained by an electron ( or proton) when moving through a potential difference of 1V
• 1eV = 1 (1.6 x 10^-19 coul)(V)
• 1 eV = 1.6 x 10^-19 Joules

• The test charge moves from an area of higher to an area of lower potential

• The negative sign indicates that V_B < V_A, or that point B is at a lower potential than at point A
• Suppose a test charge charge q_o moves from A to B
• Then, DU = q_o DV = -q_o Ed
• Therefore, if q_o is positive than the change in potential energy is negative.
• i.e.. A positive charge will lose electric potential energy when is moves in the direction of the electric field.
• On the other hand, if q_o is negative, then the change in potential energy is positive
• A negative charge gains electric potential when it moves in the direction of the electric field E.

• V_B- V_A = D V = - I ( integral from A to B) E (dot product) ds  = -I ( integral from A to B) E( cosq) ds  = -E I ( integral from A to B) ds  = -Ed   = (V_C - V_A)
• The change in potential energy for a test charge is DU = q_o DV = -q_o Ed
• Potential is a scalar quantity!
• Note that all points in a plane perpendicular to E are at the same potential
• i.e.. V_B = V_C.
• V_B and V_C are on an equipotential surface

• Since  DU = q_o DV, no work is being done when a charge is being moved along an equipotential surface

• Electric field due to Q is  E= (Qr^)/ ( 4pe_or²)
• V_B- V_A  = - I ( integral from A to B) E (dot product) ds  =  - I ( integral from A to B) [(Q)/ ( 4pe_or²)] r^(dot product) ds
• Note that r^ ( dot product) ds = (ds) cosq
• (ds) cosq can be considered as projection of ds onto r therefore (ds) cosq = dr
• Any displacement ds produces a change dr in the magnitude of r
• V_B- V_A    =  - I ( integral from A to B) [(Q)/ ( 4pe_or²)] r^(dot product) ds =  - I ( integral from A to B) [(Q dr)/ ( 4pe_or²)]
• = Q/ 4pe_or ( evaluated form rb to ra) = Q/ 4pe_o [(1/rb)- (1/ra)]
• If we take point A to be infinitely far away, then V_B = (Q/ 4pe_or )(1/rb) or similarly V = (Q/ 4pe_o)(1/r), which is the potential of a point charge.