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;

Verified step by step guidance

1. 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.
2. 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: \(-49.0 \mu C = -49.0 \times 10^{-6} C\).
3. 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.
4. 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 \(\approx 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2\).
5. 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 \(E\).

Key Concepts

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.

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.

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.