chapter24solutions.pdf

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Gauss' Law Homework Solutions

1. A cm-square in the - plane sits in a uniform electric field N C.%! B C #! $! &! E i j ka bFind the electric flux through the square.

Since the square is in the - plane, only electric field in the (perpendicular) -direction contributesB C Dto the flux. Therefore:

FI I E &! "' )! "! # $ #a b N C cm N m C

(Alternatively, you can use , where cm is the area vector.)FI "'E A A k #

2. A thundercloud produces a vertical electric field of magnitude kN C at ground level. You hold#)! a cm cm sheet of paper horizontally below the cloud.##! #)!

(a) What is the electric flux through the sheet?

FI IE #)! "! !!'"' "(# "! $ # $ #N C m N m C

(b) What would the flux be if you tilt the sheet of paper by ?$!

F )I IE "(# "! !)''! "%* "! cos a b$ # $ #N m C N m C

(c) What would the flux be if you hold the sheet of paper vertically?

Since the electric field would be parallel to the paper, the flux would be zero .

3. A cylindrical metal can has a height of cm and a radius of cm. The electric field is directed#( ""outward along the entire surface of the can (including the top and bottom), with a uniform magnitude

of N C. How much charge does the can contain?%! "! &

The area of the side of the can is m m m , the area of the top is#


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4. A -nC point charge is located at the center of a cube of side length m. What is the electric'! #!flux through eachof the faces of the cube?

By Gauss' Law, the total flux coming out of the cube is

F %I U '! "! ))& "! '() !* "# # # # C C N m N m C.

By symmetry, this flux must be evenly split between the six faces. Therefore, the flux through each

face is N m"'#a b'() G ""! N m C .#

5. A spherical ball of plasma has a radius of m. The electric field on the surface of the ball is!&!measured to be N C and points radially inward toward the center. What is the average)* "! $

volume charge density of the plasma?

By Gauss' Law, the charge inside is

U IE I% < ))& "! )* "! % !&! #%( "!% F % % 1 1! ! !# "# $ (#

I a b C.

Thus the volume charge density is

31 1

%( "! U U #%( "!

< %$ !&!volume

CC m .%

$$

(

$( $a b

a b a b

(The charge is negative since the electric field is pointing inwards.)

6. In the air over a particular region at an altitude of m above the ground, the electric field is'!!"#! '&! "!!N C directed downward. At m above the ground, the electric field is N/C downward.What is the average volume charge density in the layer of air between these two elevations?

Consider a large box of length , width , and height m, sitting so that its base is at an altitudeP [ &!of m, and its top is at an altitude of m. Then the flux out of the bottom of the box is'!! '&!

Fout IE "#! P[ a bN C

and the flux into the top is

Fin IE "!! P[ a bN C .

By Gauss' Law the total charge inside the box is

U "(( "! P[ inside out in% F F!"! #

a b C m .

The volume of the box is m , so the average volume charge density inside the box isa b&! P[

3 $& "! U "(( "! P[

&! P[ volume m

C mC m .

a ba b

"! #"# $

(The charge is positive since there is more flux out than flux in.)


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7. Charge is placed on a metallic sphere which is surrounded by air. If the radius of the sphere

is cm, how much charge can be placed on the sphere before the air near the sphere suffers electric"&breakdown? The critical electric field strength that leads to breakdown in air is N C.$! "! '

If we place a charge on the sphere, the electric field outside the sphere is governed by Coulomb'sUlaw

I " U

%


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9. An insulating sphere with a radius of cm carries a uniform volume charge density of C m .#! "& $

Find the magnitude of the electric field at a point inside the sphere that lies cm from the center.)!

A point cm from the center will only feel a force from the charge)! withina radius of cm. (The)!charge between cm and cm will not affect it) The total amount of charge within a< )! < #!radius of cm is)!

U Z "& "! !!)! $#"( "!%

$3 1 a b' $ *$C m m C

This charge has the shape of a sphere, so the electric field is determined by Coulomb's law:

I " U " $#"( "!

% < % !!)!1% 1%! !#

*

#

C

ma b%& "! $ N C

10. A square metal plate with a thickness of cm has no net charge and is placed in a region of"&uniform electric field N C directed perpendicularly to the plate. Find the resulting surface)! "! %

charge density on each face of the plate.

Since the plate is conductive, the electric field will induce positive and negative charges on opposite

sides of the plate so that the total electric field inside the plate is zero. Suppose that the surface

charge density on each side of the plate is . Then the total electric field will be5

I )! "! # #

total %! !

N C 5 5

% %

Setting this equal to zero and solving for gives5

5 % )! "! (" "! !% ( #

N C C m

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