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=−(1/4πϵ0)2qa/y2.
(d) What are the answers to parts (a) and (c) if the two charges are interchanged so that +q is at y = +a 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 −a≤y≤a (between the two charges), y > a (above both charges) and y < − a (below both charges).
(a) |y| < a:
V=kqa+y−kqa−y=2kqya2−y2
y > a:
V=kqa+y−kqy−a=−2kqay2−a2
y < a:
V=−kqa+y−kq−y+a=2kqay2−a2
A general expression valid for any y is
V=k(−q|y−a|+q|y+a|)
(b) The graph of V versus y is sketched in Figure 23.20.
(c) y >> a:V=kqa+y−kqy−a=−2kqay2−a2≈−2kqay2
(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=kqx−2kqx−a=−kq(x+a)x(x−a)
0 < x < a:
V=kqx−2kqa−x=kq(3x−a)x(x−a)
x < a:
V=−kqx+2kqx−a=kq(x+a)x(x−a)
A general expression valid for any y is
V=k(q|x|−2q|x−a|)
(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≈−kqxx2=−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#23.22Consider 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=√a2+y2.
(a) V=kq|y|−2kqr=kq(1|y|−2√a2+y2)
(b) V = 0, when y2=a2+y24
3y2=a2
y=±a√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≈kq(1y−2y)=−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.
Post a Comment for "Electric Potential Problems and Solutions 2"