University

Physics 2

Physics 2

Fall 2024

Each figure shows a closed cylinder in a uniform electric field or near some charge. Consider the electric field flux through the top of the cylinder \(\Phi_{\rm{top}}\), through the curved wall \(\Phi_{\rm{wall}}\), through the bottom \(\Phi_{\rm{bot}}\), and the total flux \(\Phi_{\rm{net}}\).

Determine if each flux is positive, negative or zero.

(Note the line and plane charges are infinite in extent.)

letter | \(\Phi_{\rm{top}}\) | \(\Phi_{\rm{wall}}\) | \(\Phi_{\rm{bot}}\) | \(\Phi_{\rm{net}}\) |
---|---|---|---|---|

a | \(+\) | \(0\) | \(-\) | \(0\) |

b | \(0\) | \(0\) | \(0\) | \(0\) |

c | \(-\) | \(-\) | \(-\) | \(-\) |

d | \(-\) | \(+\) | \(+\) | \(0\) |

e | \(+\) | \(0\) | \(-\) | \(0\) |

f | \(-\) | \(0\) | \(+\) | \(0\) |

g | \(0\) | \(+\) | \(0\) | \(+\) |

h | \(+\) | \(0\) | \(+\) | \(+\) |

i | \(0\) | \(+\) | \(0\) | \(+\) |

For this closed cylinder, \(\Phi_{\rm{top}} = -15\) N·m²/C and \(\Phi_{\rm{bot}} = 5\) N·m²/C.

What is \(\Phi_{\rm{wall}}\)?

\(\Phi_{\rm{wall}} + \Phi_{\rm{top}} + \Phi_{\rm{bot}} = 0\)

So

\(\Phi_{\rm{wall}} = 10\) N·m²/C

Find the electric flux through each of the closed surfaces. (The charges in the gray areas are inside the closed surfaces.)

Give your answers in terms of \(q\) and \(\epsilon_0\).

- \(\Phi = q / \epsilon_0\)
- \(\Phi = -q / \epsilon_0\)
- \(\Phi = 0\)

A positively charged balloon expands as it is blown up, increasing in size from initial to final diameters shown.

Do the electric fields at points 1, 2, and 3 increase, decrease, or stay the same?

**Point 1** stays the same. It starts and remains at
zero.

**Point 2** decreases. Initially there is an outward
electric field, and at the final size the field at 2 is zero.

**Point 3** stays the same. Initially there is an
outward electric field, and it remains constant as the sphere expands.
The field is the same as that due to a point charge at the center of the
sphere.

A piece of of metal has a cavity inside it. For the following cases, determine the total charge on the surface of the cavity, \(Q_c\), and the total charge on the outer surface of the metal, \(Q_o\).

- The metal holds a net charge of \(q=+4.0\) nC, and the cavity is empty.
- The metal holds zero net charge, and the cavity contains a charge of \(q=-4.0\) nC.
- The metal holds a net charge of \(q=+4.0\) nC, and the cavity contains a charge of \(q=-3.0\) nC.

- \(Q_c = 0\), \(Q_o = +4.0\) nC
- \(Q_c = +4.0\) nC, \(Q_o = -4.0\) nC,
- \(Q_c = +3.0\) nC, \(Q_o = +1.0\) nC,

- Calculate the electric flux through the open hemispherical surface due to the uniform electric field \(\vec{E} = E_0 \hat{k}\).
- If the hemisphere is rotated by \(90^{\circ}\) around the x-axis, what is the flux through it?

- Flux through the curved surface is the same as the flux through the circle at the bottom, \(\pi R^2\).

\(\Phi = E_0 \pi R^2\)

- \(\Phi = 0\)

Two large copper plates facing each other have charge densities \(\pm 4.0\) C/m² on the surface facing the other plate, and zero in between the plates. Find the electric flux through a 3 cm × 4 cm rectangular area between the plates, as shown below, for the following orientations of the area.

- If the area is parallel to the plates, and
- if the area is tilted \(\theta = 30^{\circ}\) from the parallel direction. Note, this angle can also be \(\theta = 180^{\circ} + 30^{\circ}\).

The magnitude of the electric field is

\(E = \frac{\sigma}{\epsilon_0} = 4.52 \times 10^{11}\) N/C

- \(\Phi = EA = 5.42 \times 10^{8}\) N·m²/C
- \(\Phi = EA \cos \theta = 4.70 \times 10^{8}\) N·m²/C

Shown below are two concentric conducting spherical shells of radii \(R_1\) and \(R_2\), each of finite thickness much less than either radius. The inner and outer shell carry net charges \(q_1\) and \(q_2\), respectively, where both \(q_1\) and \(q_2\) are positive.

Find the electric field in the following regions. Use \(r\) for the distance from the center, and \(\hat{r}\) for the direction radially outward from the center.

- \(r \lt R_1\)
- \(R_1 \lt r \lt R_2\)
- \(r \gt R_2\)

Find the net charge on the following surfaces.

- the inner surface of the inner shell
- the outer surface of the inner shell
- the inner surface of the outer shell
- the outer surface of the outer shell

- \(\vec{E} = \vec{0}\)
- \(\displaystyle \vec{E} = \frac{q_1}{4 \pi \epsilon_0 r^2} \hat{r}\)
- \(\displaystyle \vec{E} = \frac{q_1+q_2}{4 \pi \epsilon_0 r^2} \hat{r}\)
- \(0\)
- \(q_1\)
- \(-q_1\)
- \(q_1+q_2\)

A point charge of \(5.0 \times 10^{-8}\) C is placed at the center of an uncharged spherical conducting shell of inner radius 6.0 cm and outer radius 9.0 cm.

Find the electric field at the following distances from the center. Use \(\hat{r}\) for the direction radially outward from the center.

- \(r=4.0\) cm
- \(r=8.0\) cm
- \(r=12.0\) cm
- What are the charges induced on the inner and outer surfaces of the shell?

- \(\displaystyle \vec{E} = \frac{5.0 \times 10^{-8}}{4 \pi \epsilon_0 (0.04)^2} \hat{r} = (2.81 \times 10^{5}) \hat{r}\) N/C
- \(\vec{E} = \vec{0}\)
- \(\displaystyle \vec{E} = \frac{5.0 \times 10^{-8}}{4 \pi \epsilon_0 (0.12)^2} \hat{r} = (3.12 \times 10^{4}) \hat{r}\) N/C
- \(-5.0 \times 10^{-8}\) C and \(5.0 \times 10^{-8}\) C, respectively

Two parallel, infinite planes of charge have charge densities \(2\sigma\) and \(-\sigma\). A Gaussian cylinder with cross-section area \(A\) extends distance \(L\) to either side.

- Is \(\vec{E}\) perpendicular or parallel to the Gaussian surface at the top, bottom, and wall of the cylinder?
- Determine the electric field at the top of the cylinder \(\vec{E}_{\rm{top}}\), at at the bottom \(\vec{E}_{\rm{bot}}\).
- Find the electric fluxes through the three surfaces in terms of \(\sigma\), \(L\), \(L_0\) and \(A\). (You may not need all of these.)
- How much charge is enclosed with in the cylinder? Write \(Q_{\rm{in}}\) in terms of \(\sigma\), \(L\), \(L_0\) and \(A\).
- By combining your answers above, show that Gauss’s Law is satisfied.

The solution depends on the field from an infinite plane with surface charge density \(\sigma\):

\(\displaystyle \vec{E} = \pm \frac{\sigma}{2 \epsilon_0} \hat{n}\)

where the direction is always normally away from the plane. Note the field is uniform – it does not decrease with distance. With two planes, simply add the two resulting fields.

- top: perpendicular, bottom: perpendicular, wall: parallel
- \(\vec{E}_{\rm{top}} = \frac{\sigma}{2
\epsilon_0} \hat{k}\)

\(\vec{E}_{\rm{bot}} =-\frac{\sigma}{2 \epsilon_0} \hat{k}\) - \(\Phi_{\rm{top}} = \frac{\sigma A}{2
\epsilon_0}\)

\(\Phi_{\rm{wall}} = 0\)

\(\Phi_{\rm{bot}} = \frac{\sigma A}{2 \epsilon_0}\) - \(Q_{\rm{in}} = 2\sigma A - \sigma A = \sigma A\)
- \(\Phi_{\rm{net}} =
\Phi_{\rm{top}}+0+\Phi_{\rm{bot}} = \frac{\sigma
A}{\epsilon_0}\)

and \(Q_{\rm{in}} = \sigma A\). So

\(\Phi_{\rm{net}} = \frac{Q_{\rm{in}}}{\epsilon_0}\), which is Gauss’s Law.

*Last modified: Sun October 29 2023, 08:22 AM.*