Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Chapter 22--Examples. 2 Problem A charge, +q, is surrounded by a thin, spherical shell of radius, a, which has a charge density of –  on its surface.

Similar presentations


Presentation on theme: "1 Chapter 22--Examples. 2 Problem A charge, +q, is surrounded by a thin, spherical shell of radius, a, which has a charge density of –  on its surface."— Presentation transcript:

1 1 Chapter 22--Examples

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

3 3 Step 1: Pick your shape I choose spherical! So

4 4 Region 1: r

5 5 Region 2: a<=r<=b q enclosed =q+(4  a 2 )*(-  ) +q a b -- ++

6 6 Region 3: r>b q enclosed =q+(4  a 2 )*(-  )+ (4  b 2 )*(  ) +q a b -- ++

7 7 Problem An electric filed given by E=4i-3(y 2 +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

8 8 First, let’s get a sense of direction Planes 1. y-z: Normal to +x 2. x-z: Normal to –y 3. y-z: Normal to –x 4. x-z: Normal to +y 5. x-y: Normal to +z 6. x-y: Normal to -z X=1.0 m X=3.0 m x y z

9 9 Need to find q enclosed

10 10 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

11 11 The rest of the sides Since E is perpendicular to sides 5 & 6, the result is zero.

12 12 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. a) Prove that E = 0 for r>a b) Prove that E=0 for r>b c) Prove that, for a

13 13 First, I choose a shape I choose cylindrical! So

14 14 For r

15 15 For a

16 16 For r>b q enclosed = L- L=0 a b  This is the principle of a coaxial cable

17 17 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

18 18 First, let’s do a solid cylinder of radius, R

19 19 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’

20 20 E hole = E solid cylinder -E off-axis hole All of these are constants and do not depend on r.


Download ppt "1 Chapter 22--Examples. 2 Problem A charge, +q, is surrounded by a thin, spherical shell of radius, a, which has a charge density of –  on its surface."

Similar presentations


Ads by Google