Other Charge Distributions - Fields & Potentials

#Advanced Applications of Charge Distributions, Gauss' Law, and Electric Potential

This section dives into more complex scenarios involving charge distributions, Gauss' Law, and electric potential calculations. It builds upon the foundational concepts covered earlier, so make sure you're comfortable with those before proceeding. Let's get started! 🚀

#Extended Charge Distributions

So far, we've treated charged objects as point charges. Now, we'll tackle situations where charge is spread out over a line, ring, or sheet. The key idea is to break down the total charge Q into tiny pieces dq, each contributing a small electric field dE. Then, we integrate to find the total field E. Here's the general formula:

$$dE = k \frac{dq}{r^2} \hat{r}$$

where r is the radius vector.

#Ring of Charge Example

If x >>> a, the equation simplifies to that of a point charge, which is a great way to check your work.

#Line of Charge Example

First, express dq in terms of dy using the linear charge density λ:

Then, integrate to find the total electric field, using the Pythagorean theorem to find r:

#Gauss' Law for Various Shapes

Gauss' Law is your best friend for finding electric fields due to symmetrical charge distributions. Here's a quick rundown of how to apply it to different shapes:

#Line of Charge

Enclose the line of charge in a cylinder. Use the linear charge density λ = Q/L to find the enclosed charge and A = 2πrL for the surface area.

#Point, Hoop, or Sphere (fully enclosed)

Enclose the charge in a sphere. Use A = 4πr² for the surface area.

#Sphere (not fully enclosed)

#Insulating Sheet of Charge

Enclose the sheet in a rectangle. Remember that the field exists on both sides of the sheet. Use the area charge density σ = Q/A, so q_enc = σA and E∲dA = 2EA.

#Potential Difference for a Variety of Shapes

To find the potential difference, first use Gauss' Law to find an expression for E, then plug that into:

$$ΔV = -\int E \cdot dl$$

#Line of Charge Example

#Conducting Sheets Example

#Final Exam Focus

• Extended Charge Distributions: Be ready to calculate the electric field and potential due to lines, rings, and sheets of charge. Remember to break the charge into dq and integrate.
• Gauss' Law: Master applying Gauss' Law to various symmetrical charge distributions (spheres, cylinders, sheets). Choose the right Gaussian surface!
• Potential Difference: Know how to calculate potential differences by integrating the electric field. Pay attention to the limits of integration.
• Symmetry: Always look for symmetry in the problem. It will often simplify your calculations.
• Connections: Be prepared for questions that combine concepts from different units. For example, a question might ask you to find the electric field using Gauss' Law and then calculate the potential difference.

#Practice Questions

#

Practice Question

Multiple Choice Questions

1. 

a)

b)

Answer:

(a) The shells are conducting, so the charge is only on their surfaces. Between r1 and r2, we fully enclose Q1. Using Gauss' Law, A is the correct answer.

(b) Relative to infinity, the total potential at r2 is V = k(Q1 + Q2)/r2. Between the spheres, there's no change due to V_Q2 because the potential is constant inside a conductor. However, V_Q1 still varies according to 1/r. Thus, E is the correct answer.

2. 

Answer:

B is correct. Every point is equidistant from a given x-value, so V = kQ/r, where r = √(x² + a²).

3. 

Answer:

D is correct. Outside the spheres, E = 0 since the total enclosed charge is 0. Inside the negative shell, the potential looks like a positively charged sphere (constant when r < a, and proportional to 1/r when r > a).

#

Practice Question

Free Response Question

In the figure above, a nonconducting solid sphere of radius a with charge +Q uniformly distributed throughout its volume is concentric with a nonconducting spherical shell of inner radius 2a and outer radius 3a that has a charge -Q uniformly distributed throughout its volume. Express all answers in terms of the given quantities and fundamental constants.

(a) Using Gauss’s law, derive expressions for the magnitude of the electric field as a function of radius r in the following regions.

i. Within the solid sphere (r < a)

ii. Between the solid sphere and the spherical shell (a < r < 2a)

iii. Within the spherical shell (2a < r < 3a)

iv. Outside the spherical shell (r > 3a)

(b) What is the electric potential at the outer surface of the spherical shell (r = 3a)? Explain your reasoning.

(c) Derive an expression for the electric potential difference V_X - V_Y between points X and Y shown in the figure.

#Scoring Guidelines

(a) i. (3 points)

• Correct Gaussian surface (1 point)
• Correct enclosed charge (1 point)
• Correct electric field (1 point)

$$E(4πr^2) = \frac{1}{ε_0} \frac{Qr^3}{a^3}$$

$$E = \frac{Qr}{4πε_0a^3}$$

(a) ii. (2 points)

• Correct enclosed charge (1 point)
• Correct electric field (1 point)

$$E(4πr^2) = \frac{Q}{ε_0}$$

$$E = \frac{Q}{4πε_0r^2}$$

(a) iii. (3 points)

• Correct Gaussian surface (1 point)
• Correct enclosed charge (1 point)
• Correct electric field (1 point)

$$E(4πr^2) = \frac{1}{ε_0} (Q - Q \frac{r^3 - 8a^3}{19a^3})$$

$$E = \frac{Q}{4πε_0r^2} (1 - \frac{r^3 - 8a^3}{19a^3})$$

(a) iv. (2 points)

• Correct enclosed charge (1 point)
• Correct electric field (1 point)

$$E(4πr^2) = 0$$

$$E = 0$$

(b) (2 points)

• Correct potential (1 point)
• Correct reasoning (1 point)

The electric potential is zero because the net charge enclosed by the sphere is zero.

(c) (3 points)

• Correctly identifies the potential at point X (1 point)
• Correctly identifies the potential at point Y (1 point)
• Correct potential difference (1 point)

$$V_X = \frac{Q}{4πε_0(2.5a)}$$

$$V_Y = \frac{Q}{4πε_0(2a)}$$

$$V_X - V_Y = \frac{Q}{4πε_0} (\frac{1}{2.5a} - \frac{1}{2a})$$

$$V_X - V_Y = \frac{-Q}{20πε_0a}$$