Tuesday, January 26, 1999
Announcements:
None
Lecture notes:
Example: Electric Field Due to a Dipole along the dipole axis

• What is the electric field E at point P due to the dipole of chares +q and -q?

• The electric field due to point P is the electric field due to +q added to the electric field due to -q
• E₊ = ( 1/ 4pe_o) (q / r(+)²) and  E- = ( 1/ 4pe_o) (-q / r(-)²)
• Therefore E_total= (q/ 4pe_o) [(1/r(+)²) - (1/ r(-)²)]
• If r(+) = z -(d/2) and r(-) = z + (d/2), then E_total= (q/ 4pe_o) [(1/(z -(d/2))²) - (1/ (z + (d/2))²)].
• Further simplification leads to E_total= (q/ 4pe_oz²) [(1-(d/2z))^-2) - (1+ (d/2z))^-2)].
• Usually z >>d, so we do a Taylor expansion
• (1-(d/2z))^-2) = ( 1 + 2(d/2z) + (2(3)/2!)(d/ 2z)²) +....)  and (1+ (d/2z))^-2) =( 1 - 2(d/2z) + (2(3)/2!)(d/ 2z)²) -....)
• Therefore: E =(q/ 4pe_oz²) [ 2(d/2z) + 2(d/2z)] = (q/ 4pe_oz²) (2 d/z) = (qd/ 2pe_oz³)
• We define qd =P as the dipole moment and E =(P/ 2pe_oz³)
• P is a vector and it's direction is along the axis of the dipole, from -q to +q

• A conducting rod of length L has a charge -q uniformly distributed along its length.

• First determine how much charge there is in section dx  (dq = ldx)

 

• dx is very small compared to the total length so it can be treated as a point charge
• Contribution to electric field at P due to dq = ldx?
• dE =  (ldx)/(4pe_o(a + L -x )²)
• The electric field at P is the sum over all the contributions from different dx's along the enitre rod
• 
• 
• Complete this problem as part of the homework assignment

• The first step is to assign a coordinate system

• Charge Density = q/L =l
• The contribution to the total Electric Field due to dx
• dE = (ldx)/(4pe_or²)
• Only the y components need to be considered. Why? The x components will be equal and opposite to another x component and will therefore all cancel
• You are left with dE_y= (ldx cos q )/(4pe_or²)
• This expression is not easy to integrate becuase it contains three variables, x,r, and q.
• Everything should be put into terms of one variable, namely q.
• After solving for q (which should be done on your own time for practice) the resulting equation is

• E_y= ( 2l/ 4pe_oy) cos q dy

•  sinqm = (L/2)/ R = (L/2)/(((L/2)² + y²) ^(1/2))
• Therefore, after all substitutions and simplifications have been made E_y= (lL)/ (2pe_oy((L² + 4y²) ^(1/2)))
• What if the rod is infinitely long?