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 > the potential at a point on the positive -axis is given by V=(1/4πϵ0)2qa/y2.

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

Answer:

For a point charge, V=kqr. 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 aya (between the two charges), y > a (above both charges) and y < − (below both charges). 

(a) |y| < a:

V=kqa+ykqay=2kqya2y2

y > a:

V=kqa+ykqya=2kqay2a2

y < a:

V=kqa+ykqy+a=2kqay2a2

A general expression valid for any y is

V=k(q|ya|+q|y+a|)

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

(c) y >> a

V=kqa+ykqya=2kqay2a22kqay2

(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 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 at points on the x-axis as a function of 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=kqx2kqxa=kq(x+a)x(xa)

0 < x < a:

V=kqx2kqax=kq(3xa)x(xa)

x < a:

V=kqx+2kqxa=kq(x+a)x(xa)

A general expression valid for any y is

V=k(q|x|2q|xa|)

(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

Vkqxx2=kqx

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 at points on the y-axis as a function of 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=a2+y2.

(a) V=kq|y|2kqr=kq(1|y|2a2+y2)

(b) V = 0, when y2=a2+y24

3y2=a2

y=±a3

(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:

Vkq(1y2y)=kqy

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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