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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.

Short Answer

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

Step by step solution

01

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 \(-15.0 \times 10^{-6} \mathrm{C}\).
02

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 \(E_{ ext{inside}} = 0\).
03

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:\[ E_{\text{outside}} = \frac{8.99 \times 10^9 \cdot (-15.0 \times 10^{-6})}{(0.06)^2} \approx -3.75 \times 10^7 \mathrm{N/C} \]
04

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:\[ E = \frac{8.99 \times 10^9 \cdot (-15.0 \times 10^{-6})}{(0.11)^2} \approx -1.12 \times 10^7 \mathrm{N/C} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

22.42. A Sphere in a Sphere. A solid conducting sphere carry- ing charge \(q\) has radius a. It is inside a concentric hollow conducting sphere with inner radius \(b\) and outer radius \(c\) . The hollow sphere has no net charge. (a) Derive expressions for the electric-field magnitude in terms of the distance \(r\) from the center for the regions \(rc .\) (b) Graph the magnitude of the electric field as a function of \(r\) from \(r=0\) to \(r=2 c\) (c) What is the charge on the inner surface of the hollow sphere? (d) On the outer surface? (e) Represent the charge of the small sphere by four plus signs. Sketch the field lines of the system within a spherical volume of radius 2 c.

22\. 28 . A square insulating sheet 80.0 \(\mathrm{cm}\) on a side is held horizontally. The sheet has 7.50 \(\mathrm{nC}\) of charge spread uniformly over its area. (a) Calculate the electric field at a point 0.100 \(\mathrm{mm}\) above the center of the sheet. (b) Estimate the electric field at a point 100 \(\mathrm{m}\) above the center of the sheet. (c) Would the answers to parts (a) and (b) be different if the sheet were made of a conducting material? Why or why not?

22.22. (a) At a distance of 0.200 \(\mathrm{cm}\) from the center of a charged conducting sphere with radius \(0.100 \mathrm{cm},\) the electric field is 480 \(\mathrm{N} / \mathrm{C}\) . What is the electric field 0.600 \(\mathrm{cm}\) from the center of the sphere? (b) At a distance of 0.200 \(\mathrm{cm}\) from the axis of a very long charged conducting cylinder with radius \(0.100 \mathrm{cm},\) the electric field is 480 \(\mathrm{N} / \mathrm{C}\) . What is the electric field 0.600 \(\mathrm{cm}\) from the axis of the cylinder? (c) At a distance of 0.200 \(\mathrm{cm}\) from a large uniform sheet of charge, the electric field is 480 \(\mathrm{N} / \mathrm{C}\) . What is the electric field 1.20 \(\mathrm{cm}\) from the sheet?

22.50. (a) How many excess electrons must be distributed uni- formly within the volume of an isolated plastic sphere 30.0 \(\mathrm{cm}\) in diameter to produce an electric field of 1150 \(\mathrm{N} / \mathrm{C}\) just outside the surface of the sphere? (b) What is the electric field at a point 10.0 \(\mathrm{cm}\) outside the surface of the sphere?

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