A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 12.012.0 cm, giving it a charge of −49.0−49.0 μμC. Find the electric field just inside the paint layer.

Ch 22: Gauss' Law

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 22: Gauss' Law

Problem 9a

Chapter 22, Problem 9a

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 $negative 49.0$ $μ$ C. Find the electric field just inside the paint layer.

Verified step by step guidance

Understand that the problem involves a uniformly charged sphere, and we need to find the electric field just inside the surface of the sphere.

Recall that for a uniformly charged spherical shell, the electric field inside the shell is zero. This is due to the symmetry of the charge distribution, as per Gauss's Law.

Apply Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space: $ \Phi = \frac{Q}{\varepsilon_0} $.

Since the charge is only on the surface and we are considering a point just inside the surface, the enclosed charge is zero, leading to zero electric field inside.

Conclude that the electric field just inside the paint layer is zero, as the charges on the surface do not affect the field inside the sphere.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field

The electric field is a vector field around a charged object where a force would be exerted on other charges. It is defined as the force per unit charge and is measured in newtons per coulomb (N/C). For a spherical charge distribution, the electric field inside a conductor is zero, while outside, it behaves as if all the charge were concentrated at the center.

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It states that the total electric flux is equal to the enclosed charge divided by the permittivity of free space. This principle is particularly useful for calculating electric fields of symmetric charge distributions, such as spherical ones.

Conductors in Electrostatic Equilibrium

In electrostatic equilibrium, the electric field inside a conductor is zero, and any excess charge resides on the surface. This is because charges redistribute themselves to cancel any internal electric fields. For a charged spherical conductor, the electric field just inside the surface is zero, as charges only affect the field outside the conductor.

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Textbook Question

A flat sheet of paper of area $0.250$ m² is oriented so that the normal to the sheet is at an angle of $60$ ° to a uniform electric field of magnitude $14$ N/C. For what angle $ϕ$ between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

The nuclei of large atoms, such as uranium, with $92$ protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately $7.4$\times$10^{-15}$ m. The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

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Textbook Question

The nuclei of large atoms, such as uranium, with $92$ protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately $7.4 \times 10^{- 15}$ m. What is the electric field this nucleus produces just outside its surface?

You measure an electric field of $1.25 \times 10^{6}$ N/C at a distance of $0.150$ m from a point charge. There is no other source of electric field in the region other than this point charge.

(a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius $0.150$ m?

(b) What is the magnitude of this charge?

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Textbook Question

A hemispherical surface with radius $r$ in a region of uniform electric field $\vec{E}$ has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

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Textbook 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 $−49.0$ $μ$ C. Find the electric field just outside the paint layer;

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