A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37×10−6 C/m2. A charge of −0.500 μC is now introduced at the center of the cavity inside the sphere. (b) Calculate the strength of the electric field just outside the sphere?

Verified step by step guidance

1. Identify the total charge on the sphere by calculating the charge due to the surface charge density. Use the formula for the surface area of a sphere, \( A = 4\pi r^2 \), where \( r \) is the outer radius of the sphere. Multiply the surface area by the surface charge density to find the total charge on the sphere.
2. Recognize that the introduction of a negative charge inside the cavity induces an equal and opposite charge on the inner surface of the sphere. This leaves the outer surface with a charge equal to the original total charge of the sphere.
3. Apply Gauss's Law, \( \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \), where \( \Phi_E \) is the electric flux, \( Q_{\text{enc}} \) is the enclosed charge, and \( \epsilon_0 \) is the permittivity of free space. Consider a Gaussian surface just outside the outer surface of the sphere.
4. Since the electric field inside a conductor is zero, the electric field just outside the sphere is due only to the charge on the outer surface. Calculate the electric field using the formula \( E = \frac{Q}{4\pi \epsilon_0 r^2} \), where \( Q \) is the charge on the outer surface and \( r \) is the outer radius of the sphere.
5. Conclude that the electric field just outside the sphere is directed radially outward and has a magnitude calculated from the previous step.

Key Concepts

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 proportional to the enclosed charge, allowing us to calculate electric fields in symmetric charge distributions. In this scenario, it helps determine the electric field outside the hollow conducting sphere by considering the total charge on the sphere's surface.

Electric Field due to Surface Charge

The electric field just outside a charged conductor is determined by the surface charge density. For a uniformly charged sphere, the electric field at a point outside the sphere can be calculated using the formula E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space. This concept is crucial for finding the electric field strength just outside the conducting sphere.

Superposition Principle

The Superposition Principle states that the total electric field created by multiple charges is the vector sum of the electric fields produced by each charge individually. In this problem, the negative charge at the center of the cavity affects the distribution of charge on the conducting sphere, but the surface charge density remains uniform, allowing us to apply this principle to find the resultant electric field outside the sphere.