A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 12.012.012.0 cm, giving it a charge of −49.0−49.0−49.0 μμμC. Find the electric field just outside 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

Problema 9b

Capitolo 22, Problema 9b

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;

Guida verificata passo dopo passo

First, recognize that the problem involves a uniformly charged sphere. According to Gauss's Law, the electric field just outside a uniformly charged sphere behaves as if all the charge were concentrated at the center of the sphere.

Identify the total charge on the sphere, which is given as $-49.0 \mu C$. Convert this charge into Coulombs for standard SI unit calculations: $-49.0 \mu C = -49.0 \times 10^{-6} C$.

Determine the radius of the sphere. The diameter is given as 12.0 cm, so the radius $r$ is half of that: $r = \frac{12.0}{2} = 6.0$ cm. Convert this to meters: $r = 0.06$ m.

Apply Gauss's Law to find the electric field $E$ just outside the sphere. Gauss's Law states: $\Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0}$, where $\Phi$ is the electric flux, $Q_{enc}$ is the enclosed charge, and $\varepsilon_0$ is the permittivity of free space $\approx 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2$.

Since the electric field is uniform over the surface of the sphere, $E$ can be taken out of the integral, and the surface area $A$ of the sphere is $4\pi r^2$. Thus, $E \cdot 4\pi r^2 = \frac{Q_{enc}}{\varepsilon_0}$. Solve for $E$: $E = \frac{Q_{enc}}{4\pi \varepsilon_0 r^2}$. Substitute the known values to find $E$.

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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 just outside the surface can be calculated using Gauss's Law.

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Intro to Electric Fields

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. For a sphere, this simplifies the calculation of the electric field outside a uniformly charged surface.

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Gauss' Law

Spherical Symmetry

Spherical symmetry implies that the properties of the system are uniform in all directions from the center. In the context of a charged sphere, this means the electric field at any point just outside the sphere depends only on the radial distance from the center, simplifying calculations using Gauss's Law.

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