# CT27--5 A spherical shell with a uniform positive charge density by Ax2YAZF

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```									CT27-9. A spherical shell with a uniform positive charge density on its surface is near a
positive point charge. Is the electric field inside the sphere zero?

A) E=0 inside
+       +
+                   +
B) E  0 inside
+
+
E=?                 +     C) Not enough info to answer.
+
+                      +
+           +
+

Answer: E  0 inside the sphere. The total electric field inside the sphere is the vector
sum ofthe field due the shell of charge and the field due to the point charge:
                
E tot  Eshell  E po int . One can show from Gauss's Law that Eshell =0 (see lecture notes).
But the presence of the shell in no way affects the field due to the point charge . So
inside the shell, the total field is just the field due to the point charge.

The shell of charge must be on the surface of an insulator, not a metal. If the shell where
made of metal, then the charges on the spherical surface would rearrange in response to
the E-field of the outside point charge, and the charge distribution would not be uniform.

+       +
+                   +
+
+
+
+
+                      +
+           +
+
CT27-10.
A capacitor consists of two parallel metal plates that have been charged up with equal
and opposite charges: +Q on one plate, –Q on the other. All the excess charge resides on
the inside surfaces as shown. (Why no charge on the outside surfaces?... Because
opposite charges attract.)
The surface charge density on each plate is of magnitude . (+ on
the inside left plate, – on the inside right plate.) If the plates are large
++++++++++++++++++++++

--------------------

enough, "edge effects" are small and the magnitude of the electric field
between the plates is nearly uniform.

What is the magnitude of the electric field in the space between the
plates?
                                                   2
A)                                B)                     C)
o                            2 o                   o


o

We can see this answer is two ways:
E2
++++++++++++++++++++++

--------------------

1) The E-field outside a charged metal surface in

electrostatic equilbrium is always
o   .

2) The E-field due to a single plane of charge is E =
E1                               /(2o). In this problem, there are two planes of
charge: the left plane, plane 1, and the right plane,
plane 2. The total electric field is the vector sum of
the fields due to the two planes:
           
E tot  E1  E 2 . Every between the plates, the fields
add, so we have (between the plates)
Etot = /(2o)+ /(2o) = /o.
Everywhere outside the plates, the fields due to the two planes are in opposite directions
and cancel, Etot = 0.

```
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