An infinitely long conducting cylindrical rod of radius1.73mm is surrounded by a

Question

An infinitely long conducting cylindrical rod of radius1.73mm is surrounded by an infinitely long conducting cylindrical shell with an inner radius 3.21mm , and an outer radius 6.98mm . The figure gives you an idea of what it looks like.

     Now Gauss's law works in cases of symmetry, but there are real world complications. Therefore we are going to look at a length 3.67m of both cylinders. The inner cylinder has a charge of 2.21nC uniformly distributed along that length and the outer shell has a charge of 4.72nC uniformly distributed along that length.

Therefore we will be able to ignore the ends of the cylinders because E?  and A?  are perpendicular there.

Part B

What is the charge on the inner surface of the outer shell?

Part C

What is the charge on the outer surface of the cylindrical shell?

Explanation / Answer

Solution.

The charge on the inner surface of the shell will be zero ,because considering the gausian surface in the middle of the second cylinder. The Gausian surface lies inside the conductor. Since insde the conductor electric field is zero and therefore total charge on the inner surface of the outer shell will be zero because it lies iside the Gaussian surface.

E =0   

q = 0

Part C

Charge on the outer surface of the cylinder will be

q = linear charge density (Lenth)

q = 2lamda ( Length) Here 2lamda is the linear charge density and Length is the lenth of outer cyliner.

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