Problem 9 - University Physics with Modern Physics

Question

22.9. A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter \(12.0 \mathrm{cm},\) giving it a charge of \(-15.0 \mu \mathrm{C}\) . Find the electric field (a) just inside the paint layer; (b) just outside the paint layer: (c) 5.00 \(\mathrm{cm}\) outside the surface of the paint layer.

Step-by-Step Solution

Answer

(a) 0 N/C, (b) \(-3.75 \times 10^7\) N/C, (c) \(-1.12 \times 10^7\) N/C.

Step 1: Understand Concepts and Given Data

We are dealing with a charged spherical surface. We know the diameter of the sphere is 12.0 cm, which makes the radius 6.0 cm or 0.06 m. The charge on the sphere is \(-15.0 \mu \mathrm{C}\), which is equivalent to \(-15.0 \times 10^{-6} \mathrm{C}\).

Step 2: Just Inside the Paint Layer

According to Gauss's Law, the electric field inside a charged conducting sphere is zero as all the charge resides on the surface. Therefore, the electric field just inside the paint layer is \(E_{ ext{inside}} = 0\).

Step 3: Just Outside the Paint Layer

For the electric field just outside a uniformly charged spherical shell, we use the formula:\[ E = \frac{k \cdot Q}{r^2} \]where \(k\) is Coulomb's constant \( (8.99 \times 10^9 \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2) \), \(Q\) is the charge, and \(r\) is the radius of the sphere which is 0.06 m. Substituting the values:\[ E_{\text{outside}} = \frac{8.99 \times 10^9 \cdot (-15.0 \times 10^{-6})}{(0.06)^2} \approx -3.75 \times 10^7 \mathrm{N/C} \]

Step 4: 5.00 cm Outside the Paint Layer

Now, we find the electric field at a point 5.00 cm outside the surface. The new radius becomes \(0.06 + 0.05 = 0.11\) m. Using the formula for the electric field at a point outside the sphere:\[ E = \frac{k \cdot Q}{r^2} \]Substituting the values:\[ E = \frac{8.99 \times 10^9 \cdot (-15.0 \times 10^{-6})}{(0.11)^2} \approx -1.12 \times 10^7 \mathrm{N/C} \]

Key Concepts

Electric Field Calculation

Electric fields describe the force per unit charge exerted on a small positive test charge placed in the field. They are vital in understanding how charges interact in space. Calculating the electric field involves using Gauss's Law for symmetrical charge distributions such as spheres. For a point directly outside the surface of a charged sphere, the electric field is given by the equation:\[ E = \frac{k \cdot Q}{r^2} \]Here,

• \(E\) is the magnitude of the electric field.
• \(k\) is Coulomb's constant, a fundamental value for calculating electric fields.
• \(Q\) is the total charge on the sphere.
• \(r\) is the distance from the center to the point where we're calculating the field.

This formula stems from the idea that a charged spherical shell behaves as though all its charge is concentrated at its center. Therefore, no matter where you are in space, you can calculate the electric field at any distance by considering the entire charge at the center.

Spherical Surface Charge

A spherical surface charge refers to a distribution of electric charge spread uniformly over the surface of a sphere. This is a common assumption in physics because it makes calculations more manageable, particularly when using Gauss's Law. The charged sphere's surface is in electrostatic equilibrium, meaning once the paint (or charge) is evenly spread, the charges stay in place unless moved by an external force.
Gauss's Law is particularly useful when handling spherical charge distributions. It tells us that the electric field inside the sphere is zero. This happens because the charges reside solely on the surface, leaving the inner region field-free. When considering just outside the surface, the sphere acts like a point charge at its center, allowing us to use the same formula as we would for any point charge to find the field just outside it.

Coulomb's Constant

Coulomb’s constant \(k\) plays a crucial role in electric field equations. It is a constant that appears in Coulomb's Law, which defines the force between two point charges. Its value, approximately \(8.99 \times 10^9 \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2\), is derived from the permittivity of free space, influencing how charges interact across space.
In the realm of electric fields, Coulomb's constant helps relate the magnitude of the force between charges to the field they produce. It affects how strong the electric field is at a point around a charge, aiding in determining the force felt by other charges in proximity. Knowing this constant allows us to calculate electric fields reliably in various scenarios, including charged spherical surfaces as explained in the earlier sections.