4- +-0.62 points Tipler6 22.P053 My Notes Ask Your Teacher The figure below show

Question

4- +-0.62 points Tipler6 22.P053 My Notes Ask Your Teacher The figure below shows a portion of an infinitely ong, concentric cable in cross section. The inner conductor has a near charge densit/ of conductor has no net charge 90 n m and the outer cm 9cm 13 cm (a) Find the electric field for all values of R, where R is the perpendicular distance from the common axis of the cylindrical system. (Use the following as necessary: R answer.) Assume R is in meters and E is in N/C. Do not include units in you E(R

Explanation / Answer

Electric field inside a conductor is zero. So
E(R<0.015) = E(045<R<0.065) = 0

For no net charge on the outer cylinder, Electric field at all other positions is given by
E = 2 k (lamada) /(R)
k = 9x109
Hence E(0.015<R<0.045) = E(R>0.065)2*9*109*4.9*10-9 /R = 88.2/R

Charge per unit length, on the inner and ouer surface of outer cylinder are equal in magnitude and equal to 'lamda' given for inner conductor.
Charge density of cylinderical surface = lamada/ 2 pi R
Hence , charge density of inner surface = 4.9/ 2 *3.14*0.045 = 17.34 nC/m2
charge density of outer surface = 4.9/ 2 *3.14*0.065 = 12 nC/m2

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