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

1. The problem statement, all variables and given/known data
Someone posted a similar problem here: https://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)
 
Re: Deriving equation for electric potential along yaxis for a linear electric quadru

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

:approve:
 
Re: Deriving equation for electric potential along yaxis for a linear electric quadru

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

:approve:
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.
 
Re: Deriving equation for electric potential along yaxis for a linear electric quadru

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

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