Physics 21
Spring,
2011
Solution to HW4
235
In Figure 2329, a proton is a distance
d/
2 directly
above the center of a square of side
d
. What is the magnitude
of the electric flux through the square? (Hint: Think of the
square as one face of a cube with edge
d
.)
Electric flux through a surface is defined by the integral Φ
E
=
integraltext
E
·
d
A
. The brute force method of solving this problem
would be to write out an expression for
E
as a function
of position on the surface of the square, and integrate that
expression. But the hint tells us to use symmetry to solve
the problem.
If we imagine the proton is located in the center of a cube
with edge
d
, then it is in a position with a great deal of
symmetry: it is a distance
d/
2 from the center of
each
face
the cube. We conclude that the flux through each face of
the cube must be equal to 1
/
6 of the total flux through the
cube. But how do we calculate the total flux through the
cube?
The cube is totally enclosing the proton.
An interesting
property of flux is that the total flux through
any
enclosing
surface is the same. We saw in Lecture 3 that a sphere is
the easiest surface to calculate the total flux due to a point
charge. The result of that calculation is Φ
E
=
q
ǫ
0
.
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 Spring '08
 Hickman
 Physics, Electric charge

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