Tuesday, February 16, 1999
Announcements: None

Lecture notes:

•  V = V₁ + V₂
•  V = ( Q1/ 4pe_or1)  + ( Q2/ 4pe_or2)
• the result is a scalar quantity
• Similarly,

• V = S V_i =  S (Q_i / 4pe_or_i)
• What is the change in potential energy if we bring a charge q from point A to point B in the presence of point charge Q?
• DU = U_B - U_A = q(V_B - V_A ) = q [ Q/4pe_o][ (1/r_B) - (1/r_A)]
• Consider the case when r_A = infinity, then using the above expression it is natural to choose U_A = 0 for as the potential approaches infinity.
• U_B  = q [ Q/4pe_o] (1/r_B)
• Thus, in general....  U(r) = q [ Q/4pe_or]
• Therefore, the potential energy of the system of charges, q and Q, at a distance of r from each other is equal to the change in potential energy when we bring then from infinitely far apart to a separation of r.
• work is done by external forces
• If work is done by electrostatic forces on a charge ( or charges) then DU = - W_{due to electrostatic force}.
• In order to have DU  with no change in kinetic energy, there must be another external applied force, W_applied.
• DKE = 0 = Work done by all forces
• 0 = W_{electrostatic} + W_applied.
• DU = W_applied
• See problem 56E from the textbook

• This is similar to calculating electric field due to a charge distribution, but easier because potential is a scalar
• Example 1 : Line of charges

• What is V at P?
• The charge distribution is l = q/ L
• See page 612 in textbook for calculations
• V = [l = 4pe_o] ln [ (L +(L² + d²)^1/2)/d]

• We deduced, DV = V_f - V_i = - I (integral evaluated form i to f) E (dot product) ds
• This allows us to calculate the potential from the electric field
• The above expression also means... dV = - E (dot product) ds
• If E only has one component, E_x, then E (dot product) ds = E_x and E_x = - dV/dx
• If the charge distribution has spherical symmetry then E_r = - dV/dr.
• In general, the electric potential is a function of all three cartesian coordinates
•  V= V(x,y,z)
• Then dV =  - E (dot product) ds = -E_xdx - E_ydy - E_zdz

• From dV = - E (dot product) ds
• Assume we displace a test charge of vector [ds] that lies within a equipotential surface, then dV = o
• Therefore, electric field is perpendicular to the equipotential surface.

• E (r) = - dV/dr

• The electric field inside a conductor is zero and decreases with distance from the sphere

• The electric potential is uniform inside the sphere and decreases with distance from the sphere.