Deriving equation for electric potential along yaxis for a linear electric quadrupole

Minhtran1092

1. The problem statement, all variables and given/known data
Someone posted a similar problem here: http://www.physicsforums.com/showthread.php?t=375737

The diagram and problem description is essentially the same except I am trying to find the expression for the electric potential on the y-axis at distances y>>s.

2. Relevant equations

V = U/Q; Voltage at a distance 'r' from source charge, on a point charge Q whose potential energy is given by: k(q_source charge)(Q_point charge)(1/r)

3. The attempt at a solution
To find the voltage, I figured to take the sum of the voltage at a position 'y' from the point of origin where -2q charge was. The total electrical potential at this point was given by:
ƩV = ƩU/Q_P; ƩU = k(q)(Q_P)(1/r) + k(q)(Q_P)(1/r) + k(-2q)(Q_p)(1/r); where r was (y-s), (y+s), (y), respectively. Algebraically simplifying this sum gives: 2kq(s^2)*[1/((y^3)-y(s^2))].

For y>>s, I figured the denominator would simplify to y^3-y but that wasn't the correct answer. It seems that I had trouble understanding the general procedure for evaluating a limiting case. Mathematically, how does considering when y>>s simplify the equation to kQ/(y^3); Q = 2qs^2? (The answer is correct for the x and y axis, which makes sense intuitively since at any long distance the linear electric quadrupole can be treated as a point charge whereby the calculation to find the electrical potential on any axis is arbitrary)

issacnewton

well the potential on y axis would be

[tex]V(y)=\frac{k_e q}{(y-s)}-\frac{2k_e q}{y}+\frac{k_e q}{(y+s)} [/tex]

which simplifies to

[tex]\frac{k_e q}{y}\left [\frac{1}{(1-\frac{s}{y})}-2+ \frac{1}{(1+\frac{s}{y})}\right ][/tex]

since y >> s , we have [itex]\frac{s}{y} \ll 1 [/itex] we can use binomial expansions

[tex]\frac{1}{(1-\frac{s}{y})}\approx 1+\frac{s}{y}+\frac{s^2}{y^2} [/tex]

[tex]\frac{1}{(1+\frac{s}{y})}\approx 1-\frac{s}{y} +\frac{s^2}{y^2} [/tex]

since [itex]\frac{s}{y} \ll 1 [/itex] , we neglect higher order terms in [itex]\frac{s}{y}[/itex]

plug everything and I think you get what you are looking for

Minhtran1092

Ah I see. I wish my physics teacher told me about the binomial expansion ... it would have been very useful. Thank you, Isaac Newton (lol) for clarifying this approximation.

issacnewton

what text you are using.... I think precalculus covers binomial expansion