Line Charge and Charged Cylindrical Shell 1 2 3 4 5 1 1 1 9 An infinite line of

Question

Line Charge and Charged Cylindrical Shell

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An infinite line of charge with linear density 1 = 7.4 C/m is positioned along the axis of a thick insulating shell of inner radius a = 2.3 cm and outer radius b = 4.2 cm. The insulating shell is uniformly charged with a volume density of = -789 C/m3.

1)

What is 2, the linear charge density of the insulating shell?

C/m

2)

What is Ex(P), the value of the x-component of the electric field at point P, located a distance 8.7 cm along the y-axis from the line of charge?

N/C

3)

What is Ey(P), the value of the y-component of the electric field at point P, located a distance 8.7 cm along the y-axis from the line of charge?

N/C

4)

What is Ex(R), the value of the x-component of the electric field at point R, located a distance 1.15 cm along a line that makes an angle of 30o with the x-axis?

N/C

5)

What is Ey(R), the value of the y-component of the electric field at point R, located a distance 1.15 cm along a line that makes an angle of 30o with the x-axis?

N/C

Explanation / Answer

1) Just think of this as taking a small slice of the cylinder and finding the charge in that slice. We want to go from volume density to linear density so we need to multiply by cross-sectional area.

Area = pi*b^2 - pi*a^2

2 = *A = *pi*[b^2 - a^2] = -3.06 C/m

2) With a cylindrical symmetry, once again the E fields will only point along the radial direction. At point P the X axis is perpendicular to the radial so there’s no X component of the field.

So, Ex(P) = 0

3) Apply Gauss’s law again for the cylindrical symmetry:

  E.dA = Qenc/o =========> E = enc/(2*pi*r*o)

Since P is outside the insulating shell we need to add the total linear charge densities of both the line of charge and the shell to get our enc.

E = enc/(2*pi*r*o) = (7.4 - 3.06)/(2*pi*0.087*8.85*10^-12) = 897113.4 N/C

4) First we need to find the E field from:

E = enc/(2*pi*r*o) *cos

Keeping in mind that we only worry about the charge enclosed, i.e. the line of charge.

E = (7.4*10^-6*cos60)/(2pi*8.85*10^-12*1.15*10^-2) = 5786030.8 N/C

5) E = 1/(2*pi*r*o) *sin = (7.4*10^-6*sin60)/(2pi*8.85*10^-12*1.15*10^-2) = 10021699.4 N/C

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