Problem 84 - University Physics with Modern Physics

Question

A uniform sphere has a radius \(R\) and a total charge \(+Q,\) uniformly distributed throughout its volume. It is surrounded by a thick spherical shell carrying a total charge \(-Q,\) also uniformly distributed, and having an outer radius of \(2 R\). What is the electric field as a function of \(R ?\)

Step-by-Step Solution

Answer

Question: Determine the electric field as a function of the radius R for a uniformly charged sphere with total charge +Q and radius R, surrounded by a uniformly charged spherical shell with outer radius 2R and total charge -Q. Answer: The electric field as a function of R is given by: $$ E(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$

Step 1: Determine the electric field of a uniformly charged sphere

First, let's find the electric field of a uniformly charged sphere. To do this, we use Gauss' Law: $$ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} $$ For a sphere of radius r (with r <= R) centered at the origin, the enclosed charge is $$ Q_{inclosed} = \frac{Q}{\frac{4}{3}\pi R^3} \times \frac{4}{3}\pi r^3 $$ Then, the electric field is same for each point on the spherical surface enclosing the uniformly charged sphere, so: $$ E \times 4\pi r^2 = \frac{Q}{\frac{4}{3}\pi R^3} \times \frac{4}{3}\pi r^3 \div \epsilon_0 $$ Solving for E, we get: $$ E_{sphere}(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$

Step 2: Determine the electric field of a uniformly charged spherical shell

Now, let's find the electric field of a uniformly charged spherical shell. To do this, we again use Gauss' Law, but this time we'll use a Gaussian surface that is a sphere centered at the origin, with 2R >= r >= R. Since there is no charge enclosed by this Gaussian surface, we have: $$ \oint \vec{E} \cdot d\vec{A} = 0 $$ Then, we can say that the electric field due to the shell at the Gaussian surface is canceled by the electric field due to the inner uniformly charged sphere: $$ E_{shell}(r) = -E_{sphere}(r) $$

Step 3: Combine the electric fields

Now, we combine the electric fields from step 1 and step 2 to obtain the total electric field as a function of R: For r <= R, the electric field is only due to the charged sphere, so: $$ E(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$ For R <= r <= 2R, the electric field is due to both the sphere and the shell, so, as we derived, the electric field is: $$ E(r) = -E_{shell}(r) = -\left(-\frac{Qr}{4\pi\epsilon_0 R^3}\right) = \frac{Qr}{4\pi\epsilon_0 R^3} $$ Thus, the electric field as a function of R is: $$ E(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$

Key Concepts

Electric Field

The concept of the electric field is vital in understanding electromagnetic interactions. An electric field is a vector field around a charged object where other charges experience a force. It essentially shows how the charge will interact with its environment. The direction of the electric field is the direction a positive test charge would move if placed in the field. This is represented mathematically by

• Field strength (\( E \)) is measured in newtons per coulomb (N/C) or volts per meter (V/m)
• The electric field is calculated from the expression \[ E = \frac{F}{q} \], where \( F \) is the force experienced by a small test charge \( q \).

Using Gauss' law, one can derive the electric field for symmetric charge distributions like spheres or cylindrical shells. It's a powerful tool because it simplifies calculations for complex charge distributions, as the integral over a surface becomes manageable by symmetry.
Understanding electric fields helps us comprehend how charges interact, move, and exert forces in various configurations.

Uniformly Charged Sphere

A uniformly charged sphere models the distribution of charge throughout a spherical volume evenly. In practice, this means each part of the sphere's volume carries the same amount of charge proportionately. When using Gauss' Law to determine the electric field due to such a sphere, we consider a Gaussian surface—a hypothetical surface that mimics the shape of the object over which the electric field is calculated.

• Inside the sphere (r \( \leq \) R), the electric field \( E \) is radial and given as \( E_{sphere}(r) = \frac{Qr}{4\pi\epsilon_0 R^3} \).
• This formula shows that within the sphere, the electric field increases linearly with distance from the center, up to the sphere’s surface.

The field is pivotal for understanding phenomena like charge distribution in insulators and can explain behaviors in materials like semiconductors.

Spherical Shell

A spherical shell is another critical structure for analyzing electric fields and potential, especially when differing charges are involved. Unlike a solid sphere, a shell only has charge distributed over its surface. For a spherical shell with total charge, we denote regions as either inside or outside the shell:

• Inside the shell (r inside where charge is placed) has an electric field of zero, \( E (r) = 0 \).
• Outside, beyond the sphere, behaves similarly to a point charge with its total charge appearing to be concentrated at its center.

Gauss' Law again assists greatly in deriving these observations because it allows us to calculate fields by evaluating simpler integral forms, linking the symmetry of a spherical shell to its charge distribution. Homogeneous spherical shells thus greatly simplify the study of electrodynamics.

Enclosed Charge

The concept of enclosed charge is fundamental when applying Gauss' Law, serving as the basis for its calculations. When using a Gaussian surface, enclosed charge refers to the total charge contained within that surface. This is pertinent since Gauss' Law states that the electric flux through a closed surface is proportional to the charge enclosed.

• Calculations involve summing products of charge density, area, or volume ratios when dealing with continuous distributions.
• Inside a spherical structure, like the one in our problem, the enclosed charge varies with the volume enclosed by the Gaussian surface. That's expressed as \( Q_{enclosed} = \frac{Q}{\frac{4}{3}\pi R^3} \times \frac{4}{3}\pi r^3 \).

Understanding enclosed charge allows for the determination of electric fields around complex charge configurations, making it a cornerstone for analyzing electrostatics. This insight streamlines calculations, enabling precise descriptions of field interactions.