Question

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 \(\times\) 10\(^{-6}\) C/m\(^2\). A charge of -0.500 \(\mu\)C is now introduced at the center of the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?

Short answer

(a) 7.01 × 10⁻⁶ C/m², (b) 7.92 × 10⁵ N/C, (c) -5.65 × 10⁴ Nm²/C.

Step-by-step solution

Determine the Original Total Charge on the Surface

The original charge on the outside surface of the sphere is calculated using the surface charge density \( \sigma \) and the surface area of the outer sphere \( A \). The formula is \( Q = \sigma \cdot A \), where \( A = 4\pi r^2 \) and \( r \) is the outer radius (0.250 m). \( Q = 6.37 \times 10^{-6} \times 4 \pi (0.250)^2 \approx 5.00 \times 10^{-6} \) C.

Determine Net Charge with Inner Charge Added

With the inner charge of \(-0.500 \mu C\) introduced, by induction, an equal and opposite charge \(+0.500 \mu C\) will appear on the inner surface, causing no net charge change. Thus, the outer surface has an increased total charge by \(+0.500 \mu C\) due to charge rearrangement, making it \( 5.00 \times 10^{-6} + 0.500 \times 10^{-6} = 5.50 \times 10^{-6} \) C.

Calculate the New Surface Charge Density

Given that the charge on the outer surface is now \( 5.50 \times 10^{-6} \) C, the new surface charge density \( \sigma' = \frac{Q}{A} \). With \( A = 4 \pi (0.250)^2 \), \( \sigma' \approx \frac{5.50 \times 10^{-6}}{4 \pi (0.250)^2} \approx 7.01 \times 10^{-6} \) C/m\(^2\).

Calculate Electric Field Just Outside the Outer Surface

Using Gauss's Law, the electric field \( E \) just outside a conductive surface relates to the charge density as \( E = \frac{\sigma'}{\varepsilon_0} \), where \( \varepsilon_0 \approx 8.85 \times 10^{-12} \) C\(^2\)/(N·m\(^2\)). \( E \approx \frac{7.01 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 7.92 \times 10^{5} \) N/C.

Calculate Electric Flux Inside Inner Surface

Inside a conductor, since the electric field is zero, and considering a Gaussian surface just inside the inner surface encloses the charge \(-0.500 \mu C\), the electric flux \( \Phi \) through this surface can be calculated using \( \Phi = \frac{Q_{enc}}{\varepsilon_0} \). \( \Phi = \frac{-0.500 \times 10^{-6}}{8.85 \times 10^{-12}} \approx -5.65 \times 10^4 \) Nm\(^2\)/C.

Key concepts

Gauss's Law

Gauss's Law is a fundamental principle that helps us understand electric fields around charged objects. Imagine wrapping an imaginary closed surface, known as a Gaussian surface, around a charge or a system of charges. Gauss's Law tells us that the electric flux through this surface is directly proportional to the total charge enclosed within the surface.

This can be mathematically expressed as \( \Phi = \frac{Q_{enc}}{\varepsilon_0} \), where \( \Phi \) is the electric flux and \( \varepsilon_0 \) is the permittivity of free space. It helps in calculating electric fields for symmetrical distributions like spheres and cylinders. In this particular exercise, Gauss's Law was used to determine the electric field just outside the conducting sphere by considering the surface charge density and using the relation \( E = \frac{\sigma'}{\varepsilon_0} \). This simplifies the process of finding the electric field in situations involving symmetry, as it connects the field directly to surface characteristics.

Surface Charge Density

Surface charge density \( \sigma \) is essentially the amount of charge per unit area on a surface. It is a crucial factor when dealing with conductive materials, as it allows us to quantify how much charge resides on a surface.

For a spherical surface like the one in this exercise, the surface area can be calculated as \( A = 4 \pi r^2 \), where \( r \) is the radius of the sphere. The total surface charge \( Q \) then becomes \( Q = \sigma \cdot A \).

Initially, the outer surface charge density was given as \( 6.37 \times 10^{-6} \) C/m\(^2\), but due to the introduction of an internal charge, the surface charge density on the outer surface needed to be recalculated. With the new charge rearrangement, the updated surface charge density became \( 7.01 \times 10^{-6} \) C/m\(^2\), accounting for the influence of the inner charge on the sphere.

Electric Field Strength

Electric field strength \( E \) represents the force per unit charge exerted on a positive test charge placed in a field. It is vectorial in nature and points away from positive charges and toward negative charges.

Using Gauss's Law, one can find the electric field just outside a conductive surface through a simplified relation: \( E = \frac{\sigma'}{\varepsilon_0} \), where \( \sigma' \) is the updated surface charge density. For our hollow, conducting sphere, after including the influence of the inner charge, we calculated the electric field strength just outside the sphere as approximately \( 7.92 \times 10^{5} \) N/C.

This calculation is straightforward for symmetric charge distributions, as Gauss's Law allows us to bypass directly measuring forces and simply use field lines to depict how strong and in which direction the field acts.

Electric Flux

Electric flux \( \Phi \) describes how electric field lines "flow" through a given surface area, providing a measure of the field's intensity over that area. It's greatly employed in the context of Gauss's Law, as it relates the field over a closed surface to the charge enclosed by it.

Calculated as \( \Phi = E \times A \) when the field is uniform over the area or using Gauss's law directly when enclosed charge is considered, the electric flux offers insight into interactions in electrostatic environments.

In the exercise, the flux through a spherical surface inside the inner radius of the hollow sphere was considered. Given that the electric field inside a conductor is zero, the flux only takes into account the enclosed charge, i.e., \( \Phi = \frac{-0.500 \times 10^{-6}}{8.85 \times 10^{-12}} \), resulting in approximately \( -5.65 \times 10^4 \) Nm\(^2\)/C. This essentially visualizes how the internal negative charge modifies the sphere's immediate interior.