Electric Potential Problems and Solutions 2

  Q#23.20

A positive charge +q is located at the point x = 0, y = $-a$ and a negative charge $-q$ is located at the point x = 0, y = +a.

(a) Derive an expression for the potential at points on the -axis as a function of the coordinate Take to be zero at an infinite distance from the charges. 

(b) Graph at points on the -axis as a function of over the range from y = $-4a$ to y = +4a. 

(c) Show that for y > a the potential at a point on the positive -axis is given by $V=-(1/4\pi\epsilon_0)2qa/y^2$.

(d) What are the answers to parts (a) and (c) if the two charges are interchanged so that +q is at y = +a and $-q$ is at y = $-a$?

Answer:

For a point charge, $V=\frac{kq}{r}$. The total potential at any point is the algebraic sum of the potentials of the two charges.

Consider the distances from the point on the y-axis to each charge for the three regions $-a\leq y \leq a$ (between the two charges), y > a (above both charges) and y < − a (below both charges). 

(a) |y| < a:

$V=\frac{kq}{a+y}-\frac{kq}{a-y}=\frac{2kqy}{a^2-y^2}$

y > a:

$V=\frac{kq}{a+y}-\frac{kq}{y-a}=-\frac{2kqa}{y^2-a^2}$

y < a:

$V=\frac{-kq}{a+y}-\frac{kq}{-y+a}=\frac{2kqa}{y^2-a^2}$

A general expression valid for any y is

$V=k\left(\frac{-q}{|y-a|}+\frac{q}{|y+a|}\right)$

(b) The graph of V versus y is sketched in Figure 23.20.

(c) y >> a: 

$V=\frac{kq}{a+y}-\frac{kq}{y-a}=-\frac{2kqa}{y^2-a^2} \approx \frac{-2kqa}{y^2}$

(d) If the charges are interchanged, then the potential is of the opposite sign.

V = 0 at y = 0. V → +∞ as the positive charge is approached and V → −∞ as the negative charge is approached. 

Q# 23.21

A positive charge q is fixed at the point x = 0, y = 0 and a negative charge $-2q$ is fixed at the point x = a, y = 0.

 (a) Show the positions of the charges in a diagram. 

(b) Derive an expression for the potential V at points on the -axis as a function of the coordinate x. Take V to be zero at an infinite distance from the charges. 

(c) At which positions on the x-axis is V = 0?

(d) Graph V at points on the x-axis as a function of x in the range from x = $-2a$ to x = +2a. 

(e) What does the answer to part (b) become when x >> a? Explain why this result is obtained.

Answer:

For a point charge, $V=kq/r$. 

The total potential at any point is the algebraic sum of the potentials of the two charges. 

(a) The positions of the two charges are shown in Figure 23.21a. 

(b) x > a:

$V=\frac{kq}{x}-\frac{2kq}{x-a}=\frac{-kq(x+a)}{x(x-a)}$

0 < x < a:

$V=\frac{kq}{x}-\frac{2kq}{a-x}=\frac{kq(3x-a)}{x(x-a)}$

x < a:

$V=\frac{-kq}{x}+\frac{2kq}{x-a}=\frac{kq(x+a)}{x(x-a)}$

A general expression valid for any y is

$V=k\left(\frac{q}{|x|}-\frac{2q}{|x-a|}\right)$

(c) The potential is zero at x = −a and a/3.

(d) The graph of V versus x is sketched in Figure 23.21b.

(e) For x >> a: 

$V \approx \frac{-kqx}{x^2}=\frac{-kq}{x}$

which is the same as the potential of a point charge –q. Far from the two charges they appear to be a point charge with a charge that is the algebraic sum of their two charges. 

Q#23.22 

Consider the arrangement of point charges described in Exercise 23.21.  

(a) Derive an expression for the potential V at points on the y-axis as a function of the coordinate y. 

Take V to be zero at an infinite distance from the charges. 

(b) At which positions on the y-axis is V = 0? 

(c) Graph V at points on the y-axis as a function of y in the range from  y = $-2a$ to y = +2a.

(d) What does the answer to part (a) become when y > 0? Explain why this result is obtained.

Answer:

For a point charge, $V=kq/r$. The total potential at any point is the algebraic sum of the potentials of the two charges.

The distance of a point with coordinate y from the positive charge is |y| and the distance from the negative charge is $r=\sqrt{a^2+y^2}$.

(a) $V=\frac{kq}{|y|}-\frac{2kq}{r}=kq\left(\frac{1}{|y|}-\frac{2}{\sqrt{a^2+y^2}}\right)$

(b) V = 0, when $y^2=\frac{a^2+y^2}{4}$

$3y^2=a^2$

$y=\pm \frac{a}{\sqrt{3}}$

(c) The graph of V versus y is sketched in Figure 23.22. V → ∞ as the positive charge at the origin is approached.

(d) y >> a:

$V \approx kq\left(\frac{1}{y}-\frac{2}{y}\right)=-\frac{kq}{y}$

which is the potential of a point charge −q. Far from the two charges they appear to be a point charge with a charge that is the algebraic sum of their two charges.

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