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Chapter 21: Problem 39

A net charge of \(5.0 \mu \mathrm{C}\) is applied on one side of a solid metal sphere \(2.0 \mathrm{cm}\) in diameter. Once electrostatic equilibrium is reached, and assuming no other conductors or charges nearby, what are (a) the volume charge density inside the sphere and (b) the surface charge density on the sphere?

Short Answer

Expert verified
The volume charge density of the sphere in the state of electrostatic equilibrium is \(0 C/m^{3}\), and the surface charge density on the sphere is \(3.98 C/m^{2}\).

Step by step solution

01

Calculate Sphere Volume

The first step is to calculate the volume of the sphere using the provided diameter. The formula for the volume of a sphere is \(V = \frac{4}{3} \pi r^{3}\), where \(r\) is the radius. In this case, the diameter is \(2.0 cm\), so the radius is \(1.0 cm = 0.01 m\). Applying this to the formula gives: \(V = \frac{4}{3} \pi (0.01 m)^{3} = 4.19 \times 10^{-6} m^{3}\).
02

Find Volume Charge Density

Volume charge density (\( \rho \)) is defined as the total charge (\( Q \)) of the object divided by its volume (\( V \)). In this case, \( Q = 5.0 \mu C = 5.0 \times 10^{-6} C \), and the volume of the sphere is calculated in Step 1. However, in the state of electrostatic equilibrium, the charges in a conductor rearrange themselves and spread on the surface of the conductor, leaving no charge inside the solid metal sphere. Hence, the volume charge density in the sphere is \( \rho = 0 C/m^{3} \).
03

Find Surface Charge Density

Surface charge density (\( \sigma \)) is defined by the total charge (\( Q \)) divided by the total surface area (\( A \)). The turface area of a sphere is given by the formula \( A = 4 \pi r^{2} \), with \( r = 0.01 m \). Using this in the formula, \( A = 4 \pi (0.01 m)^{2} = 1.26 \times 10^{-3} m^{2} \). Hence, the surface charge density is \( \sigma = \frac{Q}{A} = \frac{5.0 \times 10^{-6} C}{1.26 \times 10^{-3} m^{2}} = 3.98 C/m^{2} \).

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Most popular questions from this chapter

What's the electric field strength in a region where the flux through a \(1.0 \mathrm{cm} \times 1.0 \mathrm{cm}\) flat surface is \(65 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) if the field is uniform and the surface is at right angles to the field?

A positive point charge \(q\) lies at the center of a spherical conducting shell carrying net charge \(\frac{3}{2} q .\) Sketch the field lines both inside and outside the shell, using eight field lines to represent a charge of magnitude \(q\).

A solid sphere \(10 \mathrm{cm}\) in radius carries a \(40-\mu \mathrm{C}\) charge distributed uniformly throughout its volume. It's surrounded by a concentric shell \(20 \mathrm{cm}\) in radius, also uniformly charged with \(40 \mu \mathrm{C}\). Find the electric field (a) \(5.0 \mathrm{cm},\) (b) \(15 \mathrm{cm},\) and (c) \(30 \mathrm{cm}\) from the center.

Charges \(+2 q\) and \(-q\) are near each other. Sketch some field lines for this charge distribution, using eight lines for a charge of magnitude \(q\).

Find the field produced by a uniformly charged sheet carrying \(87 \mathrm{pC} / \mathrm{m}^{2}\).
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