EEE 241 Final Exam| Questions and Answers Latest
Updated 2026/2027 (Graded A+)- ASU
1. A conducting sphere is located at the origin of the coordinate system. This sphere
has a uniform charge density such that the total charge is –Q. The radius of the
sphere is a. Concentric with the sphere is a spherical shell of radius b > a. The
spherical shell is charged to +2Q. What is the electric field for the region R > b?
Gauss’ Law:
!D• a dS = dV
n
V
4R2DR = +2Q − Q = Q
Q .
DR =
4R 2
E = DR = Q
R 0 4 0R2
2. Professor Balanis has taken a laser range finder to the roof of the pressbox at Sun
Devil stadium. It has been set up to bounce the laser light off one of the new State
Farm buildings on Rio Salado Parkway. By phase matching the reflected light
with the output of the laser source, it is found that the round trip takes the photons
exactly 2 s. What is the distance from the range finder on the roof of the
pressbox to the State Farm building?
The range is found from the condition that 2R−ct = 0 (zero phase difference in the
two waves). Hence,
ct 3 108 2 10−6
R= = = 300 m .
2 2
3. Given an electric field defined as E = xax+yay, determine the potential between
the two points P1(2,2,-1) and P2(3,1,-1).
First, it is important to note that since E has no z-component, the z-coordinates
are meaningless, especially as they are the same. Thus, our line integral is going
to be
E • dl = E x
dx + E y dy .
Because the potential is a scalar, we can chose any path we want, so we will go
along x from 2 to 3 and then along y from 2 down to 1. Thus,
3 1 3 1
x2 y2
V = − E • dl = − xdx − ydy = − − = −1 .
2 2
2 2 2 2
Updated 2026/2027 (Graded A+)- ASU
1. A conducting sphere is located at the origin of the coordinate system. This sphere
has a uniform charge density such that the total charge is –Q. The radius of the
sphere is a. Concentric with the sphere is a spherical shell of radius b > a. The
spherical shell is charged to +2Q. What is the electric field for the region R > b?
Gauss’ Law:
!D• a dS = dV
n
V
4R2DR = +2Q − Q = Q
Q .
DR =
4R 2
E = DR = Q
R 0 4 0R2
2. Professor Balanis has taken a laser range finder to the roof of the pressbox at Sun
Devil stadium. It has been set up to bounce the laser light off one of the new State
Farm buildings on Rio Salado Parkway. By phase matching the reflected light
with the output of the laser source, it is found that the round trip takes the photons
exactly 2 s. What is the distance from the range finder on the roof of the
pressbox to the State Farm building?
The range is found from the condition that 2R−ct = 0 (zero phase difference in the
two waves). Hence,
ct 3 108 2 10−6
R= = = 300 m .
2 2
3. Given an electric field defined as E = xax+yay, determine the potential between
the two points P1(2,2,-1) and P2(3,1,-1).
First, it is important to note that since E has no z-component, the z-coordinates
are meaningless, especially as they are the same. Thus, our line integral is going
to be
E • dl = E x
dx + E y dy .
Because the potential is a scalar, we can chose any path we want, so we will go
along x from 2 to 3 and then along y from 2 down to 1. Thus,
3 1 3 1
x2 y2
V = − E • dl = − xdx − ydy = − − = −1 .
2 2
2 2 2 2
