Question

You have four point charges. Their location and charges in Cartesian coordinates are: a positive charge, 2q, located at (a,0,0), another charge -2q located at (-a,0,0), a 3rd charge -q located at (0,0,b), and finally a fourth charge +5q located at (0,0,-b) (a) What is the total charge, and dipole moment, of this distribution of charges? Use the methods of "the multipole expansion" (Griffiths section 3.4.1) to find a simple approximate formula for V(r,0) (in spherical coordinates!) valid at points far from the origin. ("Simple" means only the first non-zero term is needed!) (b) Find a simple approximate expression for the electric field valid at points far from the origin. (Again, express your answer in spherical coordinates, so we want E(r, 0) Er(r, 0)f+ Eg(r,0)0 and you should figure out what Er and Ee are.) (c) Sketch this (approximate) E field. (Don't worry about what happens near the origin, I just want a sketch of the simple approximation)

Answer 1

Let q_1 = 2q r^vector'_1 = ax^^+ 0y^^+ 0z^^� q_2 = -2q r^vector'_2 = -ax^^+ oy^^+ oz^^q_3 = -q r^vector'_3 = 0x^^+ 0y^^+ bz^^� q_4 = 5q r^vector'_4 = 0x^^+ 0y^^- bz^^� Total charge Q = q_1 + q_2 + q_3 + q_4� Q = 2q - 2q - q + 5q� Q = 4q� Dipole moment of the charge configuration is P^vector = q_1 r^vector'_1 + q_2 r^vector'_2 + q_3 r^vector'_3 + q_4 r^vector'_4� p^vector = 2qax^^- 2q(-ax^^) - qbz^^- 5qbz^^� p^vector = 4qax^^- 6qbz^^� since total charge Q notequalto 0, the first \r\n non zero term in the multiple expansion is the monopole term� V almostequalto 1/4 pi epsilon_0 Q/r = 1/4 pi epsilon_0 4q/r� The electric field is related to the electric potential as� E^vector = -nabla^vector v� E^vector = -1/4 pi epsilon_0 times 4q nabla^vector(1/r)� E^vector = -4q/4 pi epsilon_0 -r^^/r^2� E^vector = 1/4 pi epsilon_0 4q/r^2

Note that in the simple approximation their is no dependency on the polar angle $\theta$. If we would got total charge to be zero, we would have moved to the dipole approximation (in case it was non-zero). The electric field and electric potential will be dependent on the polar angle $\theta$.