Question

Charge \(Q\) is distributed uniformly throughout the volume of an insulating sphere of radius \(R =\) 4.00 cm. At a distance of \(r =\) 8.00 cm from the center of the sphere, the electric field due to the charge distribution has magnitude \(E =\) 940 N/C. What are (a) the volume charge density for the sphere and (b) the electric field at a distance of 2.00 cm from the sphere's center?

Short answer

(a) \(\rho \approx 6.26 \times 10^{-4}\, \text{C/m}^3\); (b) \(E \approx 187.19\,\text{N/C}\) at 2 cm.

Step-by-step solution

Understanding the given information

We are given:- Total charge on the sphere, \(Q\) (not directly provided, we need to find it).- Radius of the sphere, \(R = 4.00\) cm.- The electric field at a distance \(r = 8.00\) cm outside the sphere, \(E = 940\) N/C.- We need to find the volume charge density \(\rho\) and the electric field at a distance \(r = 2.00\) cm.

Using Gauss's Law for points outside the sphere

For a point outside a uniformly charged sphere, the sphere can be treated as a point charge. The electric field \(E\) at a distance \(r = 8.00\) cm is given by:\[E = \frac{kQ}{r^2},\]where \(k = 8.99 \times 10^9\, \text{N} \cdot \text{m}^2/\text{C}^2\) is the Coulomb's constant.Solving for \(Q\), we have:\[Q = \frac{Er^2}{k} = \frac{940 \times (0.08)^2}{8.99 \times 10^9}.\]

Calculate the Charge Q

Substitute the values into the equation:\[Q = \frac{940 \times 0.0064}{8.99 \times 10^9} = \frac{6.016}{8.99 \times 10^7}.\]Calculate \(Q\) to find:\[Q \approx 6.69 \times 10^{-8}\, \text{C}.\]

Finding the volume charge density \(\rho\)

The volume charge density \(\rho\) is uniform throughout the sphere and is given by:\[\rho = \frac{Q}{\text{Volume of sphere}}.\]The volume \(V\) of the sphere is \(\frac{4}{3}\pi R^3\), where \(R = 0.04\) m.\[\rho = \frac{6.69 \times 10^{-8}}{\frac{4}{3}\pi (0.04)^3}.\]

Calculate \(\rho\) and find its value

Substituting in:\[\rho = \frac{6.69 \times 10^{-8}}{\frac{4}{3} \pi \times (0.064 \times 10^{-3})} \approx \frac{6.69 \times 10^{-8}}{1.07 \times 10^{-4}}.\]This gives:\[\rho \approx 6.26 \times 10^{-4}\, \text{C/m}^3.\]

Calculate the electric field inside the sphere at r = 2.00 cm

Using Gauss's law inside the sphere, the electric field \(E\) at distance \(r\) is given by:\[E = \frac{kQr}{R^3}.\]Substitute \(Q\) and calculate:\[E = \frac{8.99 \times 10^9 \times 6.69 \times 10^{-8} \times 0.02}{(0.04)^3}.\]

Find the electric field at r = 2.00 cm

Compute the value:\[E = \frac{8.99 \times 10^9 \times 6.69 \times 10^{-8} \times 0.02}{6.4 \times 10^{-5}} = \frac{1.198 \times 10^{-2}}{6.4 \times 10^{-5}}.\]Calculate:\[E \approx 187.19\,\text{N/C}.\]

Key concepts

Volume Charge Density

Volume charge density, denoted as \( \rho \), is a measure of how much electric charge is contained within a given volume of space, such as inside a sphere. It is expressed in units of coulombs per cubic meter (C/m\(^3\)).
For a uniformly charged body, like the insulating sphere in this problem, volume charge density can be determined using the formula: - \( \rho = \frac{Q}{V} \) - where \( Q \) is the total charge and \( V \) is the volume of the sphere.The volume of a sphere is calculated using the formula: - \( V = \frac{4}{3} \pi R^3 \)- where \( R \) is the radius.In this exercise, we found that the sphere's total charge was approximately \( 6.69 \times 10^{-8} \) C. After calculating the volume using the sphere's radius, you can use it to find \( \rho \). Understanding how \( \rho \) is derived helps you comprehend how electric fields arise from distributed charges.

Gauss's Law

Gauss's Law is a key principle for understanding electric fields, particularly in symmetric situations like spheres or cylinders. It states that the electric flux through a closed surface is proportional to the charge enclosed within that surface. In mathematical form, Gauss's Law is written as:- \( \Phi_E = \frac{Q_{enclosed}}{\varepsilon_0} \)- where \( \Phi_E \) is the electric flux and \( \varepsilon_0 \) is the vacuum permittivity.For a spherical charge distribution, if a point lies outside the sphere, the sphere can be treated as if all its charge were concentrated at its center. This allows for simplifications when calculating the electric field.In this problem:- At 8 cm, the electric field was calculated using the formula \( E = \frac{kQ}{r^2} \) since the point is outside the sphere.- At 2 cm, the point is inside, and the formula \( E = \frac{kQr}{R^3} \) is used, exploiting the symmetry and linear relationship of radius inside the sphere. Gauss's Law simplifies the process of calculating electric fields, especially when dealing with symmetrical charge distributions.

Coulomb's Constant

Coulomb's constant, denoted as \( k \), plays a crucial role in the equations governing electric forces and fields. Its value is \( 8.99 \times 10^9 \) N·m\(^2\)/C\(^2\), and it appears in Coulomb's Law, which describes the force between two point charges:- \( F = k \frac{|q_1q_2|}{r^2} \)- where \( F \) is the force, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.In this exercise, Coulomb's constant is used to express the electric field due to a charge:- \( E = \frac{kQ}{r^2} \) for a distance outside the sphere.This constant is essential for converting the relationship between electric field (E) and the charge causing it. It ensures the units work out correctly and provides a standard way of calculating electric forces and fields across various problems. By understanding how \( k \) fits into these calculations, you gain deeper insight into the nature of electric interactions and force.