Step-by-Step Physics Study Guidance for Midterm Exam 1

Q1. Two concentric spheres: Find the total excess charge on the inner and outer surfaces of the conducting sphere.

Background

Topic: Electrostatics, Conductors, and Charge Distribution

This question tests your understanding of how charges distribute themselves on conductors and the application of Gauss's Law to spherical geometries.

Key Terms and Formulas:

• Gauss's Law: $\Phi_E = \frac{Q_{encl}}{\varepsilon_0}$

• Conductors in electrostatic equilibrium: Excess charge resides on the surface.

• Charge conservation: The net charge on the shell is the sum of charges on its inner and outer surfaces.

Step-by-Step Guidance

1. Identify the charges: The inner sphere has $+5.00\ \mu\text{C}$, and the outer shell has $-8.00\ \mu\text{C}$.

2. Recall that the inner surface of the conducting shell must have a charge that cancels the field from the inner sphere inside the conductor.

3. Use charge conservation: The total charge on the shell is the sum of the charges on its inner and outer surfaces.

4. Set up equations for the inner and outer surface charges based on the above principles.

Try solving on your own before revealing the answer!

Final Answer:

Inner surface charge: $-5.00\ \mu\text{C}$ Outer surface charge: $-3.00\ \mu\text{C}$

The inner surface must have $-5.00\ \mu\text{C}$ to cancel the field from the inner sphere, and the outer surface has the remainder to make the total shell charge $-8.00\ \mu\text{C}$.

Q2. Find the magnitude and direction of the electric field at the center (r = 0 cm) of the inner sphere.

Background

Topic: Electric Field Inside a Charged Sphere

This question tests your understanding of electric fields inside conductors and nonconductors, especially at the center.

Key Terms and Formulas:

• Gauss's Law: $\Phi_E = \frac{Q_{encl}}{\varepsilon_0}$

• Electric field inside a solid nonconducting sphere with surface charge: $E = 0$ at the center.

Step-by-Step Guidance

1. Consider the symmetry: At the center, all contributions from surface charge cancel out.

2. Recall that for a solid nonconducting sphere with surface charge, the electric field inside is zero.

3. Use Gauss's Law to confirm the result.

Try solving on your own before revealing the answer!

Final Answer:

The electric field at the center is $0$ N/C.

By symmetry and Gauss's Law, the field at the center of a sphere with surface charge is zero.

Q3. Find the magnitude and direction of the electric field at r = 15.0 cm from the center of the inner sphere.

Background

Topic: Electric Field Between Spherical Shells

This question tests your ability to use Gauss's Law to find the electric field at a point between two concentric spheres.

Key Terms and Formulas:

• Gauss's Law: $E \cdot 4\pi r^2 = \frac{Q_{encl}}{\varepsilon_0}$

• For $r$ between $R_1$ and $R_2$, only the inner sphere's charge is enclosed.

Step-by-Step Guidance

1. Identify the region: $r = 15.0$ cm is between $R_1$ and $R_2$.

2. Apply Gauss's Law for a spherical surface at $r = 15.0$ cm.

3. Enclosed charge is only the inner sphere's charge: $Q_{encl} = +5.00\ \mu\text{C}$.

4. Set up the formula: $E = \frac{Q_{encl}}{4\pi\varepsilon_0 r^2}$.

Try solving on your own before revealing the answer!

Final Answer:

$E = \frac{5.00 \times 10^{-6}}{4\pi \times 8.85 \times 10^{-12} \times (0.15)^2}$ N/C, directed radially outward.

The field is due to the inner sphere's charge only, since $r$ is between $R_1$ and $R_2$.

Q4. Find the magnitude and direction of the electric field at r = 27.0 cm from the center of the inner sphere.

Background

Topic: Electric Field Outside a Spherical Shell

This question tests your understanding of how to use Gauss's Law to find the electric field outside the inner sphere but inside the outer shell.

Key Terms and Formulas:

• Gauss's Law: $E \cdot 4\pi r^2 = \frac{Q_{encl}}{\varepsilon_0}$

• For $r$ between $R_2$ and $R_3$, the enclosed charge includes the inner sphere and the inner surface of the shell.

Step-by-Step Guidance

1. Identify the region: $r = 27.0$ cm is between $R_2$ and $R_3$.

2. Determine the total enclosed charge: sum the inner sphere's charge and the inner surface charge of the shell.

3. Apply Gauss's Law for a spherical surface at $r = 27.0$ cm.

4. Set up the formula: $E = \frac{Q_{encl}}{4\pi\varepsilon_0 r^2}$.

Try solving on your own before revealing the answer!

Final Answer:

$E = \frac{0}{4\pi\varepsilon_0 r^2} = 0$ N/C

The field is zero because the total enclosed charge is zero (inner sphere and shell's inner surface charges cancel).

Q5. Find the magnitude and direction of the electric field at r = 35.0 cm from the center of the inner sphere.

Background

Topic: Electric Field Outside Both Spheres

This question tests your ability to use Gauss's Law to find the electric field outside the entire system.

Key Terms and Formulas:

• Gauss's Law: $E \cdot 4\pi r^2 = \frac{Q_{encl}}{\varepsilon_0}$

• For $r > R_3$, the enclosed charge is the sum of all charges.

Step-by-Step Guidance

1. Identify the region: $r = 35.0$ cm is outside both spheres.

2. Sum all charges: inner sphere and shell.

3. Apply Gauss's Law for a spherical surface at $r = 35.0$ cm.

4. Set up the formula: $E = \frac{Q_{total}}{4\pi\varepsilon_0 r^2}$.

Try solving on your own before revealing the answer!

Final Answer:

$E = \frac{-3.00 \times 10^{-6}}{4\pi \times 8.85 \times 10^{-12} \times (0.35)^2}$ N/C, directed radially inward.

The field is due to the net charge of the system, which is $+5.00\ \mu\text{C} + (-8.00\ \mu\text{C}) = -3.00\ \mu\text{C}$.