Figure shows two cocentric cylinders This shape is connected

Figure shows two co-centric cylinders. This shape is connected to a 5 volts battery. The radius of the inner cylinder is 1 mm. The radius of the outer cylinder is 2 mm. The conductivity of the inner cylinder is sigma_i =2 S/m. The conductivity of the outer section is sigma_o = 5 S/m. The height is 10 mm. Calculate the electric field in the inner cylinder. Calculate the current density in the inner cylinder Calculate the total current in the inner cylinder Calculate the resistance of the shape

Solution

As a third example, let’s consider a spherical capacitor which consists of two concentric spherical shells of radii a and b, as shown in Figure 5.2.5. The inner shell has a charge +Q uniformly distributed over its surface, and the outer shell an equal but opposite charge –Q. What is the capacitance of this configuration? Figure 5.2.5 (a) spherical capacitor with two concentric spherical shells of radii a and b. (b) Gaussian surface for calculating the electric field. Solution: The electric field is non-vanishing only in the region a r < < b . Using Gauss’s law, we obtain ( ) 2 0 4 r r S Q d E A E r = = = E A JG JG w

(5.2.8) or 2 1 4 r o

Q E r = (5.2.9)

Therefore, the potential difference between the two conducting shells is:

2 0 0 0 1 1 4 4 4 b b b a r a a Q dr Q Q b a V V V E dr r a b

= = = = = ab (5.2.10

) which yields 0 4 | | Q C V b = = ab a (5.2.11)

Again, the capacitance C depends only on the physical dimensions, a and b. An “isolated” conductor (with the second conductor placed at infinity) also has a capacitance. In the limit whereb , the above equation becomes 5-8 0 0 lim lim 4 lim 4 4 1 b b b ab a C a b a a b 0 =

= = R

(5.2.12) Thus, for a single isolated spherical conductor of radius R, the

capacitance is 0 C = 4

(5.2.13) The above expression can also be obtained by noting that a conducting sphere of radius R with a charge Q uniformly distributed over its surface has 0 V Q= / 4 R , using infinity as the reference point having zero potential,V () = 0 . This gives 0 0 4 | | / 4 Q Q C V Q R R = = =

(5.2.14) As expected, the capacitance of an isolated charged sphere only depends on its geometry, namely, the radius R.

C = 4R

C = Q / V

=Q/Q/4 R

=4 R