MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2013
Problem Set 4
Due: Tuesday, March 5 at 9 pm.
Hand in your problem set in your section slot in the boxes outside the door of 32-082 or 26-152
depending on which is your classroom. Make sure you clearly write your name, section, and
table number on your problem set.
Text: Dourmashkin, Belcher, and Liao; Introduction to E & M MIT 8.02 Course Notes
Revised.
Week Four: Equipotentials
and Energy; Exam 1
W04D01 M/T Feb 26/27
Reading
Potential and Gauss’s Law; Equipotential Lines and
Electric Fields
Course Notes: Sections 3.3-3.4, 4.4-4.6, 4.10.5
W04D2 W/R Feb 27/28
Exam 1 Review
Exam 1 Thursday Feb 28
7:30 pm –9:30 pm
W04D3 F Mar 1
No Class
Problem 1
Two charges lie on the x-axis. A representation of the equipotentials of the electric potential of
these two charges is shown to the right. The “streaks” in this representation are parallel to the
equipotential curves. Which one of the following statements is true?
1. The two charges have opposite signs and the charge on the left is smaller in magnitude
than the charge on the right.
2. The two charges have opposite signs and the charge on the left is larger in magnitude
than the charge on the right.
3. The two charges have the same sign and the charge on the left is smaller in magnitude
than the charge on the right.
4. The two charges have the same sign and the charge on the left is larger in magnitude than
the charge on the right.
Problem 2 Partial Derivatives and the Gradient
Introduction: For one-dimensional functions of a single variable, e.g. f (x) , the derivative,
df / dx or f '(x) , tells you how the function changes for small changes in the independent
variable x. When a function depends on several variables, e.g. f (x, y, z) , there are several
different derivatives, called partial derivatives, ¶f / ¶x , ¶f / ¶y , and ¶f / ¶z . Both in concept
and in practice the partial derivative is the same as the one-dimensional derivative, asking how
the function changes as you change one of its independent variables (while holding the others
fixed – treat them as constants when you take the derivative).
Consider a point-like positively charged object with charge +q located at the origin. Choose
V (¥) = 0 then we determined in class (W03D2 class) that the electric potential difference
V (r) - V (¥) = V (r) between infinity and any point on a sphere of radius r centered on the
origin is given by the expression
V (r) =
ke q
.
r
(a) Find an expression for the potential in Cartesian coordinates (x, y, z) .
(b) Calculate the three derivatives, called partial derivatives, ¶V / ¶x , ¶V / ¶y , and ¶V / ¶z .
Gradient: It is also convenient to define the gradient of a multidimensional scalar function. The
gradient is a vector field that everywhere points in the direction of maximum change (steepest
ascent uphill) of the function. It is calculated as follows:
Ñf =
(c) Again consider V (x, y, z) =
¶f
¶f
¶f
î + ĵ + k̂ .
¶x
¶y
¶z
ke q
x2 + y2 + z 2
. Calculate its gradient ÑV . Use the fact that
r = xî + yĵ + zk̂ and r = x 2 + y 2 + z 2 to express your answer in terms of ke , q , r , and r .
(d) Based on your answer to part (c), what is the relationship between ÑV and the electric field
E.
(e) (i) For a point-like charged object, does the gradient ÑV point in the direction of maximum
increase or maximum decrease of the function V ? Explain your reasoning. (ii) Does the sign of
the charge effect your answer to part (i)? Explain your reasoning.
Generalization: Finding the Electric Field from the Electric Potential
Your results for the point-like charged object generalize to any source of electric field. Because
the electrostatic force on a charged object in an electrostatic field is a conservative force, the line
B
integral
òF
elec
× d s is path independent and hence in class (W03D2) we defined the electric
A
potential difference by
B
B
Ft
× d s = - ò Es × d s .
q
A t
A
V (B) - V ( A) = - ò
Whenever a line integral of a function is independent of the path, the line integral can be written
as
B
V (B) - V ( A) = ò ÑV × d s .
A
Comparing our two expressions we see that
E = -ÑV .
(f) Using the result that E = -ÑV , does the electric field depend on your choice of point for the
zero reference electric potential? Explain your reasoning.
Problem 3 Suppose that the electric potential varies along the x-axis as shown in the figure
below. The potential does not vary in the y- or z -direction. Of the intervals shown (ignore the
behavior at the end points of the intervals), determine the intervals in which Ex has (a) its
greatest absolute value, (b) its least. (c) Plot Ex as a function of x. (d) What sort of charge
distributions would produce these kinds of changes in the potential? Where are they located?
Problem 4
Two parallel infinite non-conducting plates lying in the xy-plane are separated by a distance d .
The upper plate located at z = d / 2 is uniformly positively charged with surface charge density
+s . The lower plate located at z = -d / 2 is uniformly negatively charged with surface charge
density -s .
a) Find the electric field in the regions (i) z > d / 2 , (ii) z < -d / 2 , (iii) -d / 2 < z < d / 2 .
b) Find the electric potential difference V (d / 2) - V (-d / 2) .
Problem 5
Consider a uniformly charged sphere of radius R and charge Q . Find the electric potential
difference V (r) - V (0) between any point lying on a sphere of radius r and the point at the
origin, for (a) 0 < r < R and (b) r > R . Choose the zero reference point for the potential at the
origin, i.e. V (0) = 0 . (c) Make a plot of V (r) vs. r .