The figure below is a section of a conducting rod of radius R 1 = 1.40 mm and le

Question

The figure below is a section of a conducting rod of radiusR1 = 1.40 mm and length L = 11.80 minside a thin-walled coaxial conducting cylindrical shell of radiusR2 = 11.3R1 and the (same)length L. The net charge on the rod isQ1 = +3.65 × 10-12 C; that onthe shell is Q2 = -2.11Q1.What are the (a) magnitude E and(b) direction (radially inward (type 1) or outward(type 0)) of the electric field at radial distance r =2.24R2? What are (c)Eand (d) the direction (1 for inward, 0 foroutward) at r = 5.26R1? What is thecharge on the (e) interior and(f) exterior surface of the shell?

Give electric fields in N/C and charges in Coulombs.

Explanation / Answer

Given :   for simplicity let us assume the charge density ofboth the conducting cylinder and the shell are uniform and weneglect the    firing effect, symmetry can be used to showthat the electric field is radial both between the cylinderand the shell and    outside the shell, it is zero of courseinside the cylinder and inside the shell (a)    we take gaussian surface to be acylinder of length L coaxial with the given given cylindersand of larger radius r than    either of them    the flux through this surface is     = 2 r L E    where E is the magnitude of the field at thegaussian surface    we can ignore any flux through theends    the charge enclosed by the gaussian surfaceis    qenc = Q1 +Q2            =- Q1            =- 3.65 x 10-12 C    so the gauss law yields to    2 r o L E =qenc    E = qenc / 2 o L R       = ........ N / C    |E| = ........... N / C b ) Inward (c)       similar manner of part (a) d ) outward (e)    we consider a cylindrical gaussian surfacewhose radius places it with the shell itself the electric field iszero at all points on    the surface since any field within aconducting material would lead to current flow so the totalelectric flux through the    gaussian surface is zero and the net chargewith in it is zero since the central rod has a charge Q1the inner surface of the    shell nmust have chharge    Qin = - Q1          = -......... C   (f)    since the shell is known to have totalcharge           Q2 = -2.11Q1    it must have charge    Qout = Q2 - Qin            = - Q1             =........ C on its outer surface   Solve it    firing effect, symmetry can be used to showthat the electric field is radial both between the cylinderand the shell and    outside the shell, it is zero of courseinside the cylinder and inside the shell (a)    we take gaussian surface to be acylinder of length L coaxial with the given given cylindersand of larger radius r than    either of them    the flux through this surface is     = 2 r L E    where E is the magnitude of the field at thegaussian surface    we can ignore any flux through theends    the charge enclosed by the gaussian surfaceis    qenc = Q1 +Q2            =- Q1            =- 3.65 x 10-12 C    so the gauss law yields to    2 r o L E =qenc    E = qenc / 2 o L R       = ........ N / C    |E| = ........... N / C b ) Inward (c)       similar manner of part (a) d ) outward (e)    we consider a cylindrical gaussian surfacewhose radius places it with the shell itself the electric field iszero at all points on    the surface since any field within aconducting material would lead to current flow so the totalelectric flux through the    gaussian surface is zero and the net chargewith in it is zero since the central rod has a charge Q1the inner surface of the    shell nmust have chharge    Qin = - Q1          = -......... C   (f)    since the shell is known to have totalcharge           Q2 = -2.11Q1    it must have charge    Qout = Q2 - Qin            = - Q1             =........ C on its outer surface   Solve it          = -......... C   (f)    since the shell is known to have totalcharge           Q2 = -2.11Q1    it must have charge    Qout = Q2 - Qin            = - Q1             =........ C on its outer surface   Solve it    it must have charge    Qout = Q2 - Qin            = - Q1             =........ C on its outer surface   Solve it

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