Download Name: Electric Flux and Gauss`s Law – Practice I. Gauss: Johann

Name: _______________________________
Electric Flux and Gauss’s Law – Practice
I. Gauss: Johann Carl Friedrich Gauss (April 30, 1777 – February 23, 1855) was a German
mathematician and scientist who had a remarkable influence in many fields of mathematics and science.
He is considered to be one of history's most influential mathematicians. He was a child prodigy with
many stories about his astounding precocity as a toddler. Gauss made his first ground-breaking
mathematical discoveries while still a teenager.
II. Concept of Flux: Flux is often used in the context of a fluid
flowing through a pipe. The flux,  (the capital Greek letter phi),
is then the volume per second [m3/s]. This flux is equal to the
cross-sectional area [m2] perpendicular to the flow times the
velocity [m/s]. In general, the flux through the surface is given by
the equation:
= Av cos = A  v
where A is the area vector whose magnitude is equal to the area of
the surface and whose direction is normal to the surface. The
area, velocity and the angle between the two will affect the flux.
III. Electric Flux: The electric flux is a measure of the amount
of electric field passing through a surface. It is defined as the
product of the electric field perpendicular to the surface and the area of the surface, E  EA . In
general, the total electric is found by adding up all the contributions from many small individual areas.
E 

d E 

E  dA
IV. Gauss' Law: Gauss' Law is usually used to determine the
___________________ in situations where there is some symmetry to
the charge distribution, making the integral easier to solve.
E 

Qenclosed
E  dA 
o
Qenclosed
o
1. The square surface measures 3.2 mm on each side. It is immersed in
a uniform electric field with magnitude E = 1800 N/C and with field
lines at an angle of θ = 35° with a normal to the surface, as shown.
Take that normal to be directed “outward,” as though the surface were
one face of a box. Calculate the electric flux through the surface.
2. A point charge of 1.8 μC is at the center of a Gaussian cube 55 cm on edge. What is the net
electric flux through the surface?
3. The electric field in a certain region of Earth's atmosphere is directed vertically down. At an
altitude of 300 m, the field has magnitude 60.0 N/C, and at an altitude of 200 m, the magnitude is 100
N/C. Find the net amount of charge contained in a cube 100 m on edge, with horizontal faces at
altitudes of 200 and 300 m.
4. A uniformly charged conducting sphere of 1.2 m diameter has a surface charge density of 8.1 μC/m2.
A. Find the net charge on the sphere.
B. What is the total electric flux leaving the surface of the sphere?
Solutions:
1. -1.5 x 10-2 N m2/C
2. 2.0 x 105 N m2/C
3. 3.54 x 10-6 C
4. a. 3.7 x 10-5 C
b. 4.1 x 106 N m2/C
5. Examples Of Using Gauss' Law:
A. Point Charge:
+Q
B. Line Charge: Long straight wire
+ + + + + + + + + + + + + + +
A long, thin-walled metal tube of radius R = 3.00 cm, with a charge per unit length of  = 2.00 ×
10-8 C/m.
a. What is the magnitude E of the electric field at radial
distance r = R/2.00?
b. What is the magnitude E of the electric field at radial
distance r = 2.00R?
c. Graph E versus r for the range r = 0 to 2.00R.
Solutions: a. E = 0
b. 5.99 x 103 N/C
C. Thin Conducting Sheet of Charge:
A small, nonconducting ball of mass m = 1.0 mg and charge q = 2.0 × 10-8 C
(distributed uniformly through its volume) hangs from an insulating thread that
makes an angle θ = 30° with a vertical, thin, uniformly charged nonconducting
sheet (shown in cross section). Considering the gravitational force on the ball
and assuming the sheet extends far vertically and into and out of the page,
calculate the surface charge density σ of the sheet.
Solution: 5.0 x 10-9 C/m2
D. Thick Conducting Sheet of Charge:
E. A Nonconducting Sphere With Uniform Charge Density
Q
R
dV
Q
F. A Solid Conducting Sphere
R
dV
A solid nonconducting sphere of radius a = 2.00 cm is concentric with
a spherical conducting shell of inner radius b = 2.00a and outer radius
c = 2.40a. The sphere has a net uniform charge q1 = +5.00 fC and the
shell has a net charge q2 = -q1. What is the magnitude of the electric
field at radial distances?
a. r = 0
b. r = a/2.00
c. r = 1.50a
d. r = 2.30a
e. r = 3.50a
f. What is the net charge on the inner and outer surface of the shell?
Solutions:
a. E = 0
b. 5.62 x 10-2 N/C
c. 4.99 x 10-2 N/C
d. E = 0
e. E = 0
f. qinner = -5.00 fC, qouter = 0