Question

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 12.0 cm, giving it a charge of -49.0 \(\mu\)C. Find the electric field (a) just inside the paint layer; (b) just outside the paint layer; (c) 5.00 cm outside the surface of the paint layer.

Short answer

(a) 0 N/C; (b) -12.2 kN/C; (c) -3.64 kN/C.

Step-by-step solution

Understanding the Problem

The problem involves calculating the electric field at different points relative to a uniformly charged sphere. The charge is uniformly spread over the surface of the sphere, and we are tasked with finding the electric field at three specific locations.

Concept of Electric Field Inside a Conductor

Inside a conductor, the electric field is zero due to charges residing on the surface. Since the charged paint layer acts like a conductor, the electric field inside the paint layer is zero, using Gauss's law for conductors.

Calculate the Electric Field Just Outside the Paint

To find the electric field just outside the charged sphere, use the formula for the electric field from a charged sphere:\[ E = \frac{k \cdot Q}{r^2} \]where \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \), \( Q = -49 \times 10^{-6} \text{ C} \), and the radius \( r = 6.0 \text{ cm} = 0.06 \text{ m} \).Substitute these values into the formula to find \( E \).

Compute the Electric Field 5.00 cm Outside the Surface

For this, the distance from the center of the sphere is the sum of the sphere's radius and the additional 5.00 cm, giving a total distance \( r = 0.06 + 0.05 = 0.11 \text{ m} \). Use the same formula for the electric field:\[ E = \frac{k \cdot Q}{r^2} \]Substitute the values \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \), \( Q = -49 \times 10^{-6} \text{ C} \), and \( r = 0.11 \text{ m} \) to calculate \( E \).

Summarize the Calculations

(a) The electric field just inside the paint layer is 0 N/C. (b) Just outside, the field is calculated using the sphere's surface dimensions, resulting in a certain N/C value. (c) 5.00 cm outside the surface, the field is recalculated with the new radius, yielding a different N/C value.

Key concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism. It relates the electric field surrounding a closed surface to the charge enclosed by that surface. In mathematical terms, it is expressed as: \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]\(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is a differential area on the closed surface, \(Q_{\text{enc}}\) is the total charge enclosed, and \(\varepsilon_0\) is the vacuum permittivity.
In the context of the exercise provided, Gauss's law helps explain why the electric field inside a conductor is zero. A charged thin layer like the paint on a sphere can be imagined as a conducting surface. According to Gauss's law, any excess charge resides on the surface of a conductor, resulting in no electric field inside it. Therefore, the electric field just inside the paint layer is zero. This is a pivotal concept as it leads to understanding how electric fields behave around different geometries of charged objects.

Charged Sphere

Understanding a charged sphere is essential when calculating electric fields around it. In this exercise, the sphere is plastic with a uniform layer of charged paint on its surface. The charge on the sphere is given as \(-49.0 \mu\text{C}\). Such a configuration means that all charge resides on the outer surface of the sphere.
The sphere behaves as a point charge for field calculations outside its surface. Similar to a point charge, the field around a sphere decreases as you move further away. The symmetry of the sphere allows for simplified calculations using Gauss's law. The field within the material of the sphere (if considered conducting due to the paint) is zero, but this changes immediately outside the sphere's surface where the entire charge influences the field calculations.

Electric Field Calculations

Electric field calculations involve using known values and formulas to determine the field's strength at various points around a charged body. The formula for a charged sphere is: \[ E = \frac{k \cdot Q}{r^2} \]where \( E \) is the electric field, \( k \) is Coulomb's constant \( (8.99 \times 10^9 \text{ Nm}^2/\text{C}^2) \), \( Q \) is the charge, and \( r \) is the distance from the sphere's center.
For the paint-covered sphere, different points calculate differently. Just outside the sphere, the radius \( r \) equals the sphere's radius (0.06 m), using this radius helps calculate the electric field directly at the surface. If you're interested in a position further out, for instance, 5.00 cm beyond the sphere, you add this distance to the sphere's radius to obtain a new \( r \) (0.11 m in this case), then reapply the formula. Observing these calculations allows you to differentiate how distance impacts the intensity of the electric field surrounding charged objects.