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Chapter 22--Examples

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Problem A charge, +q, is surrounded by a thin, spherical shell of radius, a, which has a charge density of –s on its surface. This shell is, in turn, surrounded by another thin shell of radius, b, which has a surface charge of +s. Find the electric fields in Region 1: r<a Region 2: a<= r <=b Region 3: r> b +q a b -s +s

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Step 1: Pick your shape I choose spherical! So

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Region 1: r<a +q a b -s +s qenclosed=q

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**Region 2: a<=r<=b**

+q a b -s +s qenclosed=q+(4pa2)*(-s)

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Region 3: r>b +q a b -s +s qenclosed=q+(4pa2)*(-s)+ (4pb2)*(s)

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Problem An electric filed given by E=4i-3(y2+2)j pierces the Gaussian cube shown below. (E is in newtons/coulomb and y is meters). What net charge is enclosed by the Gaussian cube? X=1.0 m X=3.0 m x y z

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**First, let’s get a sense of direction**

4 y Planes y-z: Normal to +x x-z: Normal to –y y-z: Normal to –x x-z: Normal to +y x-y: Normal to +z x-y: Normal to -z 6 1 3 x 2 z 5 X=3.0 m X=1.0 m

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Need to find qenclosed

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**Integrating each side (start with surface 1)**

Region 3, in which the normal vector points in the opposite direction, will have a value of -16

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The rest of the sides Since E is perpendicular to sides 5 & 6, the result is zero.

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Problem The figure below shows a cross-section of two thin concentric cylinders with radii of a and b where b>a. The cylinders equal and opposite charges per unit length of l. Prove that E = 0 for r>a Prove that E=0 for r>b Prove that, for a<r<b, a b -l l

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First, I choose a shape I choose cylindrical! So

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For r<a qenclosed =0 a b -l l

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For a<r<b qenclosed =lL a b -l l

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**For r>b qenclosed =lL-lL=0 This is the principle of a coaxial cable**

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Problem A very long, solid insulating cylinder with radius R has a cylindrical hole with radius, a, bored along its entire length. The axis of the hole is a distance b from the axis of the cylinder, where a<b<R. The solid material of the cylinder has a uniform charge density, p. Find the magnitude and direction of the electric field inside the hole and show that E is uniform over the entire hole. R b a

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**First, let’s do a solid cylinder of radius, R**

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**Now what if we have an off-axis cylinder**

We learned in Phys 250, that we can “translate” coordinates by r’=r-b Where b is the direction and distance of the center of the off-axis cylinder r is a vector from the origin b r’ r

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**Ehole= Esolid cylinder-Eoff-axis hole**

All of these are constants and do not depend on r.

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Electric Field Models The electric field of a point charge q at the origin, r = 0, is where є 0 = 8.85 × 10 –12 C 2 /N m 2 is the permittivity constant.

Electric Field Models The electric field of a point charge q at the origin, r = 0, is where є 0 = 8.85 × 10 –12 C 2 /N m 2 is the permittivity constant.

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