Find Voltage on Y-Axis for y>>s

[Minhtran1092]

Homework Statement

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.

Homework 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)

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

$$V(y)=\frac{k_e q}{(y-s)}-\frac{2k_e q}{y}+\frac{k_e q}{(y+s)}$$

which simplifies to

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

since y >> s , we have $\frac{s}{y} \ll 1$ we can use binomial expansions

$$\frac{1}{(1-\frac{s}{y})}\approx 1+\frac{s}{y}+\frac{s^2}{y^2}$$

$$\frac{1}{(1+\frac{s}{y})}\approx 1-\frac{s}{y} +\frac{s^2}{y^2}$$

since $\frac{s}{y} \ll 1$ , we neglect higher order terms in $\frac{s}{y}$

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

 

[Minhtran1092]

well the potential on y-axis would be

$$V(y)=\frac{k_e q}{(y-s)}-\frac{2k_e q}{y}+\frac{k_e q}{(y+s)}$$

which simplifies to

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

since y >> s , we have $\frac{s}{y} \ll 1$ we can use binomial expansions

$$\frac{1}{(1-\frac{s}{y})}\approx 1+\frac{s}{y}+\frac{s^2}{y^2}$$

$$\frac{1}{(1+\frac{s}{y})}\approx 1-\frac{s}{y} +\frac{s^2}{y^2}$$

since $\frac{s}{y} \ll 1$ , we neglect higher order terms in $\frac{s}{y}$

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

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

 

[asteriatic]

Your attempt at a solution is close, but there are a few errors. Firstly, the total electrical potential at a point is not the sum of the individual potentials at that point. It is the sum of the individual potentials at that point due to all the charges present. So the correct expression for the total potential at a point y would be:

V = k(q)(Q_P)(1/(y-s)) + k(q)(Q_P)(1/(y+s)) + k(-2q)(Q_p)(1/y)

Secondly, when considering the limiting case of y>>s, we can simplify the equation by neglecting the terms involving s in the denominator. This is because when y is much larger than s, the term y-s (or y+s) becomes approximately equal to y, and the terms involving s become negligible compared to the terms involving y. So the simplified expression would be:

V = k(q)(Q_P)(1/y) + k(q)(Q_P)(1/y) + k(-2q)(Q_p)(1/y)

= 2kqQ_P/y + (-2kq^2)/y

= 2kq(Q_P - q)/y

= kQ/y

where Q = 2qs^2.