# MyOpenMath/Solutions/Electric Flux

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### 1

Each surface of the rectangular box shown is aligned with the xyz coordinate system. Two surfaces occupy identical rectangles in the planes x=0 and x=x_{1}=3 m. The other four surfaces are rectangles in y=y_{0}=1 m, y=y_{1}=5 m, z=z_{0}=1 m, and z=z_{1}=3 m. The surfaces in the yz plane each have area 8m^{2}. Those in the xy plane have area 12m^{2} ,and those in the zx plane have area 6m^{2}. An electric field of magnitude 10 N/C has components in the y and z directions and is directed at 30° above the xy-plane (i.e. above the y axis.) What is the magnitude (absolute value) of the electric flux through a surface aligned parallel to the xz plane?^{[1]}

**Solution**: Φ=5.196E+01 N·m^{2}/C. The face is marked at the bottom of the square, and the angle of the electric field is shown as θ In this case, the flux is (10 N/C)(6m^{2})(cos30°)=**51.96 Nm**^{2}/C

### 2

Each surface of the rectangular box shown is aligned with the xyz coordinate system. Two surfaces occupy identical rectangles in the planes x=0 and x=x_{1}=3 m. The other four surfaces are rectangles in y=y_{0}=1 m, y=y_{1}=5 m, z=z_{0}=1 m, and z=z_{1}=3 m. The surfaces in the yz plane each have area 8m^{2}. Those in the xy plane have area 12m^{2} ,and those in the zx plane have area 6m^{2}. An electric field of magnitude 10 N/C has components in the y and z directions and is directed at 60° from the z-axis. What is the magnitude (absolute value) of the electric flux through a surface aligned parallel to the xz plane? ^{[2]}

**Solution:**Φ=5.196E+01 N·m2/CThis is like the previous problem, except now the angle is measured with respect to the z axis so we instead use sin60° and get the same answer: (10 N/C)(6m^{2})(sin60°)=**51.96 Nm**^{2}/C

### 3

Each surface of the rectangular box shown is aligned with the xyz coordinate system. Two surfaces occupy identical rectangles in the planes x=0 and x=x_{1}=3 m. The other four surfaces are rectangles in y=y_{0}=1 m, y=y_{1}=5 m, z=z_{0}=1 m, and z=z_{1}=3 m. The surfaces in the yz plane each have area 8m^{2}. Those in the xy plane have area 12m^{2} ,and those in the zx plane have area 6m^{2}. An electric field has the xyz components (0, 8.7, 5.0) N/C. What is the magnitude (absolute value) of the electric flux through a surface aligned parallel to the xz plane?^{[3]}

**Solution**This is the same geometry, except now we take only the y-component: Φ=(8.7N/C)(6m^{2})=**52.2 Nm**^{2}/C

### 4

What is the magnetude (absolute value) of the electric flux through a rectangle that occupies the z=0 plane with corners at (x,y)= (x=0, y=0), (x=3, y=0), (x=0, y=2), and (x=3, y=2), where x and y are measured in meters. The electric field is

**Solution:**The dimensions of the rectangle are , where and . We only care about the z component of the electric field. Instead of being the flux, we have to break the rectangle up into little rectangles, each with area , where is very small:- =
**24**(units are either V·m or N·m^{2}/C)

### 5

Five concentric spherical shells have radius of exactly (1m, 2m, 3m, 4m, 5m).Each is uniformly charged with 5 nano-Coulombs. What is the magnitude of the electric field at a distance of 3.5 m from the center of the shells?^{[4]}

**Solution:**By Gauss's law the charge enclosed by a sphere of radius 3.5 meters is determined by the fact that charged spheres have radius less than 2.5 meters:

- Spherical symmetry (or more precisely spherical invariance) renders the surface integral trivial:

- ,
- where . Equating this integral to yields the desired result:
**1.102E+01 N/C**.

### 6

A non-conducting sphere of radius R=2 m has a non-uniform charge density that varies with the distnce from its center as given by ρ(r)=ar^{2} (r≤R) where a=1 nC·m^{-1}. What is the magnitude of the electric field at a distance of 1 m from the center?^{[5]}

**Solution:**Here, , is the easy integral in Gauss's law. The difficult integral is finding the enclosed charge:- Here, is one of many notations used to denote a small volume differential (other notations being ) For our purposes, we can use symmetry to divide our sphere into small shells of volume,
- Since the variable has already been taken to denote the radius of the Gaussian sphere, we make a "change of variables",
- (using .)
- Hence the charge enclosed by a sphere of radius r is:
- I think this yeilds =
**2.259E+01 N/C**

## OER Attribution

- ↑ Inspired by Example 6.3 from OpenStax University Physics2: https://cnx.org/contents/eg-XcBxE@9.8:7Rx6Svvy@4/61-Electric-Flux_1
- ↑ Example 6.3a from OpenStax University Physics2: https://cnx.org/contents/eg-XcBxE@9.8:7Rx6Svvy@4/61-Electric-Flux_1
- ↑ Example 6.3b from OpenStax University Physics2: https://cnx.org/contents/eg-XcBxE@9.8:7Rx6Svvy@4/61-Electric-Flux_1
- ↑ Inspired by Example 6.6 from OpenStax University Physics2, but modified by user:Guy vandegrift.
- ↑ Example 6.7 from OpenStax University Physics2: https://cnx.org/contents/eg-XcBxE@9.8:7NEpGtkt@4/63-Applying-Gausss-Law_1