Physics 9 Fall 2011 Homework 4 - Solutions Friday September 16 ...

5. When an uncharged conducting sphere of radius a is placed at the origin of an xyz
coordinate system that lies in an initially uniform electric field E = E0 ˆ k, the resulting
electrostatic potential is V (x, y, z) = V0 for points inside the sphere, and
V (x, y, z) = V0 − E0z +
E0a 3 z
(x 2 + y 2 + z 2 ) 3/2
for points outside the sphere, where V0 is the (constant) electrostatic potential on the
conductor. Use this equation to determine the x, y, and z components of the resulting
electric field. What is the full electric field in vector notation?
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Solution
The electric field may be found from the potential by taking the gradient,
such that the components are Ex = − ∂V
need to take the partial derivatives.
E = −∇V,
So, the x and y components are straightforward,
Ex = − ∂V
∂x = −E0a 3 z ∂
1
∂x
and
Ey = − ∂V
∂y = −E0a 3 z ∂
∂y
∂x , Ey = − ∂V
∂y , and Ez = − ∂V
∂z
(x 2 + y 2 + z 2 ) 3/2
1
(x 2 + y 2 + z 2 ) 3/2
The z component is more work, but not too much more.
Ez = − ∂V
∂
= −
∂z ∂z
V0 − E0z +
E0a3
z
= E0−
(x 2 + y 2 + z 2 ) 3/2
=
=
3E0a 3 zx
(x2 + y2 + z2 , 5/2
)
3E0a 3 zy
(x2 + y2 + z2 . 5/2
)
E0a 3
(x 2 + y 2 + z 2 )
. Thus, we just
3/2 +
3E0a3z2 (x2 + y2 + z2 . 5/2
)
This is all of the components, and we can write the full electric field in component
form as
E = −∇V = − ∂V ∂V
î −
∂x ∂y ˆj − ∂V
∂z ˆ k,
such that
E =
3E0a 3 zx
(x 2 + y 2 + z 2 )
5/2 î+
3E0a 3 zy
(x2 + y2 + z2 ˆj+ 5/2
)
6
E0 −
E0a 3
(x2 + y2 + z2 + 3/2
)
3E0a 3 z 2
(x 2 + y 2 + z 2 ) 5/2
ˆk.