9. Consider the two long, cylindrical conductor, separated by a distance d form a capacitor, Cylinder 1 has charge density A and radius bị and Cylinder 2 has charge density -A and radius b2. Show that the capacitance per unit length is given approximately by C = T€0 ( In where b is the geometric mean 10. Using the Gauss's Law to prove the following (a) Any excess charge placed on a conductor on its surface (b) A closed, hollow conductor shields its interior from field due to the charges outside but does not shield its exterior from the fields due to charges placed inside it.

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Answer numbeer 10 

9. Consider the two long, cylindrical conductor, separated by a distance d form
a capacitor, Cylinder 1 has charge density A and radius bị and Cylinder 2 has
charge density -A and radius b2. Show that the capacitance per unit length is
given approximately by
C = T€0 ( In
where b is the geometric mean
10. Using the Gauss's Law to prove the following
(a) Any excess charge placed on a conductor on its surface
(b) A closed, hollow conductor shields its interior from field due to the charges
outside but does not shield its exterior from the fields due to charges placed
inside it.
Transcribed Image Text:9. Consider the two long, cylindrical conductor, separated by a distance d form a capacitor, Cylinder 1 has charge density A and radius bị and Cylinder 2 has charge density -A and radius b2. Show that the capacitance per unit length is given approximately by C = T€0 ( In where b is the geometric mean 10. Using the Gauss's Law to prove the following (a) Any excess charge placed on a conductor on its surface (b) A closed, hollow conductor shields its interior from field due to the charges outside but does not shield its exterior from the fields due to charges placed inside it.
Expert Solution
Step 1

According to Gauss law, the expression for electric flux is 

ϕ=E.dS=qenclosedε0

where ϕ is the electric flux, E is the electric field, qenclosed is the charge enclosed within the surface and ε0 is the permittivity of free space.

(a) Inside the conductor, electric field is zero. 

From the Gauss law expression, for conductor 

ϕ=0.dS=qenclosedε00=qenclosedε0qenclosed=0

There is no charge enclosed inside the conductor, all the charge goes to the surface of the conductor. 

 

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