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1
22A – Gauss’ Law Concepts
Calculating Flux
- Calculate the flux of an electric field of magnitude 3 N/C passing through a flat surface of area 2m
2
when the surface normal is at an angle of :
0 degrees with respect to the field___________
20 degrees with respect to the field__________
45 degrees with respect to the field__________
60 degrees with respect to the field__________
90 degrees with respect to the field__________
- Consider a right angle prism as sketched to the right. The apex
angle is = 3 0 degrees. The height is L , the width is W and the
base is B. A constant electric field E 0 passes normally into the
left face of the prism. Calculate the flux through each face of
the prism and show that the sum of the flux through ALL faces
of the prism is zero. (Give your answers in terms of L, W, and
E 0.
Φ 𝑏𝑜𝑡𝑡𝑜𝑚
= _____________ Φ 𝑙𝑒𝑓𝑡 𝑓𝑎𝑐𝑒
= __________________ Φ 𝑟𝑖𝑔ℎ𝑡 𝑓𝑎𝑐𝑒
= ___________
Φ 𝑏𝑎𝑐𝑘
= ____________________ Φ 𝑓𝑟𝑜𝑛𝑡
= ____________________
Φ 𝑡𝑜𝑡𝑎𝑙
= ________________
Background
Gauss’ Law relates the charge enclosed in a volume to the net electric field flux that passes
through the surface that encloses the volume. The double integral with the circle means one is
integrating over a closed surface.
∯ 𝐸 ⃗ ∙ 𝑑𝐴 =
𝑞 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑
𝜀 0
Eq. 22a
Gauss’ Law can be applied to find charge enclosed from the field or to find field from charge
enclosed. In order to determine the electric field at a point using Gauss’ Law it is usually
necessary for the problem to have a simple symmetry (spherical, cylindrical…). Note the vector
dot product in the integral.
B
A- 2 m ²
E- 3 ¥. 2 m? c oso ◦ = 6. O N. c m?
E: 3 £. 2 m² ✗ co sz o
E = 5. 63 8 N' c m ²
E - 3 4 m 2 m² • C o s 4 5 = 4. 24 26 μm?
E - 3 ¥- 2 m². Co s 60 = 3. 0 NÉ M²
E - 3 ¥ 2 m ². C o s 9 0 : ON - m ³ a
0 N .cm?
63 8 N 'm²
2 426 Né m²
N. m²
c
J N. c m?
C o s 1 80 :
- E IW
- I
0 + 0 + 0 + 1 - E , LW / + HE ,
I w)
Calculating Charge Enclosed from Flux
- Given an electric field of 𝑬 ⃗⃗ = (10𝑖̂ + 20𝑦𝑗̂ + 30𝑘 ̂ )(
𝑁
𝐶
)
in the region of the cube to the right, you will solve for
the charge enclosed. (Yes, the y is intended.) Distances
are in meters.
The cube is shown here on x-y-z axes. The front surface
of the cube is in the y-z plane.
a. Draw a normal unit vector in the center of the right
hand surface.
b. What is the Cartesian unit vector that represents the
normal to the right hand surface? (±𝑖̂, ±𝑗̂ , ±𝑘
̂ )
Normal vector=________
c. Write the normal vector(±𝑖̂, ±𝑗̂ , ±𝑘
̂ ) for the
left_____, bottom____, top_____, front_____, and back_____ surfaces of the cube..
Now solve for the flux through the surfaces of the cube.
d. Find the flux Φ = ∯ 𝐸 ⃗ ∙ 𝑑𝐴 for the left hand ( y = 0 ) surface. Φ 𝑦= 0 =________
f. Find the flux Φ = (^) ∯ 𝐸 ⃗ ∙ 𝑑𝐴 for the right hand ( y = L ) surface. Φ 𝑦=𝐿 =________
g. Find the flux Φ = ∯ 𝐸 ⃗ ∙ 𝑑𝐴 for the bottom ( z = 0) surface. Φ 𝑧= 0 =________
h. Find the flux Φ = (^) ∯ 𝐸 ⃗ ∙ 𝑑𝐴 through the top ( z = L) surface. Φ 𝑧=𝐿
=________
i. Find the flux Φ = ∯ 𝐸 ⃗ ∙ 𝑑𝐴 through the front ( x = 0) surface. Φ 𝑥= 0 =___________
j. Find the flux Φ = (^) ∯ 𝐸 ⃗ ∙ 𝑑𝐴 through the back ( x = - L) surface. Φ 𝑥=−𝐿
=___________
Now find the net flux through the cube surface and the resultant charge enclosed.
k. Find the total flux through all of the surfaces of the cube. Φ 𝑇𝑜𝑡𝑎𝑙 =_______________
l. Using the total flux, find the charge enclosed by the cube surfaces. 𝑞 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 =_______________
J
ñ I i (^) ↑
E. dÑ = ( 1 07 + 20 4 5 + 30 414 )
∅y : g= o
∅- y- 2 =( 1 0 5 + 2 0 95 + 306 ) ( 1 - 3 )
= 2 0 23
⇒ " ( 1 0 72 04 5 + 30 K ) (-[ E)^2 02
( 10 7 20 45 + 30 K ) ( ( 2 )
= 30 2 2 mm² / 0
301 mm² /c
0 × - 01 0 to ys + 30 k ) ( - L: )
= - 1 0 EN ink
∅ =x =e - 1 0 to ys + 30 k) ( L - i)
= 1 01 2 N m ²/⑥ 10 1 ²Nm^ ²^ /C
= 2 023 N m/ C 202 3 Nm /C
1. 7 70 8 4 × 10 - 1 °C
The arguments from question 1) can be repeated to find (^) ∮ 𝐸 ⃗ ∙ 𝑑𝐴 for r > R. In this case
the Gaussian surface encloses the total charge density.
g. Using 𝑞 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 = ∫ 𝜌(𝑟
′ ) 4 𝜋𝑟
′ 𝑑𝑟
′
𝑅
0
, find the total charge Q of the sphere.
𝑄 𝑡𝑜𝑡𝑎𝑙 =
h. Solve for E ( r ) for r > R in terms of r, R , and a.
E(r) =
2 ) Gauss’ Law can be used to find the field for other simple symmetries by choosing a Gaussian surface
appropriate to each charge distribution. IN this question you will solve for the field for a line of charge.
a) For an infinitely long, uniform, straight cylinder of charge of charge density , the appropriate
Gaussian surface is:
A) A sphere centered on the center of the line.
B) A circular cylinder coaxial with the line.
C) A cube. Answer: a). ___________
Explain your choice:
b) Assume a Gaussian surface that is a cylinder of radius y and length L centered on the
line. It can be argued that the field from the line must point radially outward from
the line, perpendicular to the line.
i. The electric flux through the top and bottom surfaces is zero. Explain why.
ii. Simplify the expression (^) ∬ 𝐸 ⃗ ∙ 𝑑𝐴 for the curved cylindrical when it is
assumed that 𝐸 ⃗ is normal to the surface. (Write your answer in terms of r, L, and E(r).)
iii. Write the charge ,𝑞 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 , enclosed in the cylinder in terms of 𝜆 𝑎𝑛𝑑 𝐿.
iv. Write the electric field at distance r from the line in terms of 𝜆, 𝑟, and constants.
𝐸(𝑦)^ =
% n⇔ B³
r
+ T. GR?
B
T he e d ge o f the cy l inde r is ra di a lly
eq ua di st a nt fr o m its c o rre s p on din g po i nt o
f (^) ineo tcha rg e
c h a nge
B ec au se th e e l ec tri c f ie ld i s pe rp en di cula r
t o th e a r ea vec tor
E: E o
{enc los ed : ✗ L
L
2 TE ,
- Gauss’ Law can also be used to calculate the field due to an infinitely large diameter sheet with
uniform charge density. Write down the equation for the electric field for an infinite flat sheet of positive
area charge density and identify the terms. (You can get this equation from the textbook or class notes!
Identify your source.)
E =
Terms:
“Good Enough for Most Applications”
- Gauss Law is only approximate for finite lines and sheets. In this section you will compare the results
of Gauss’ Law, calculated for ideal symmetries, with exact calculations using Coulomb’s Law. (If the
difference between Gauss’ Law and exact calculation is less than 2%, it’s “good enough for Physics 2 and
MasteringPhysics.”)
a. Circular Disk: The electric field on the axis of a circular disk of charge density 𝜎 is 𝐸 𝑧
(𝑧) =
𝜎
2 𝜀 0
[ 1 −
1
√(
𝑅
𝑧
)
2
] at a distance of z from the disk. Gauss’ Law predicts a field of 𝐸 𝑧 (𝑧) =
𝜎
2 𝜀 0
. At what
value of z/R does the prediction of Gauss’ Law differ from Coulomb’s law by 2%? Explain your logic in
finding this number.
b. Straight Line: In activity 21D we dragged you through the calculation of the electric field a distance r
from center of a line of charge of length L. The result was. 𝐸 =
𝜆
2 𝜋𝑟𝜀 𝑜
1
√( 4
𝑟 2
𝐿 2
. The prediction for an
infinite line is 𝐸 =
𝜆
2 𝜋𝑟𝜖 0
. At what value of r/L does the prediction of Gauss’ Law differ from Coulomb’s
law by 2%? Explain your logic in finding this number.
29 g
( N o te s )
Eo - El ect ri c per mit tiv it y
So v gr ea te r. is a s i mil a ri t y wi t h
t he 2 e qu at io ns wi th t heir fi rs t t erm th e sec o nd te r m
ca n e qu al 0. 98 ( 2 % cha n ge ). T he g r e at er t he v alu e
o f = /R Util l in t ur n l ea d to a l o we r!
t her e
Eg a u s s - Ec u l om b = 2 % = 1 - EE :L, = 3 0 = 1 - 1 % 12 {a E ga u s s
d 41 1 T H, 0. 0 2 = - 0 10 2 = I
4 2 f t.
4 7 72 21 1 = 41 17 2 + 1 ) = #. 2 = £ 98 = [I? 1 : ⅓? = 1. 04 - 1 = 0. 04 2 = 0 = ° " '
¼ - Fo l - o. I
th e clo s er t he d en o m in a to r c an be to 1 the sm al ler the
% dif fe re nc e of the 2 e qu a t ion s
..
le ss
