Problem 15 - University Physics with Modern Physics

Question

22.15. A point charge of \(+5.00 \mu C\) is located on the \(x\) -axis at \(x=4.00 \mathrm{m},\) next to a spherical surface of radius 3.00 \(\mathrm{m}\) centered at the origin. (a) Calculate the magnitude of the electric field at \(x=3.00 \mathrm{m} .\) (b) Calculate the magnitude of the electric field at \(x=-3.00 \mathrm{m} .\) (c) According to Gauss's law, the net flux through the sphere is zero because it contains no charge. Yet the field due to the external charge is much stronger on the near side of the sphere (i.e., at \(x=3.00 \mathrm{m}\) ) than on the far side (at \(x=-3.00 \mathrm{m}\) ). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)? Explain. A sketch will help.

Step-by-Step Solution

Answer

(a) 44950 N/C, (b) 918.3 N/C, (c) Flux is zero due to area and field density balance.

Step 1: Understand the Problem

We have a point charge of \(+5.00 \mu C\) located at \(x = 4.00 \, \text{m}\) and want to find the electric field at \(x = 3.00 \, \text{m}\) and at \(x = -3.00 \, \text{m}\) on a spherical surface centered at the origin with a radius of \(3.00 \, \text{m}\). We must then explain how the net flux through the sphere is zero even with field strength differences.

Step 2: Calculate Electric Field at x=3.00 m

Using Coulomb's law, the electric field \(E\) due to a point charge \(q\) at a distance \(r\) is given by:\[E = \frac{k \cdot q}{r^2}\]where \(k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\). The distance \(r\) from the charge to \(x=3.00 \, \text{m}\) is \(4.00 \, \text{m} - 3.00 \, \text{m} = 1.00 \, \text{m}\). Thus, the electric field is:\[E = \frac{(8.99 \times 10^9) \cdot (5.00 \times 10^{-6})}{1.00^2} = 4.495 \times 10^4 \, \text{N/C}\]

Step 3: Calculate Electric Field at x=-3.00 m

For \(x=-3.00 \, \text{m}\), the distance \(r\) from the charge is \(4.00 \, \text{m} + 3.00 \, \text{m} = 7.00 \, \text{m}\). Using Coulomb's law:\[E = \frac{(8.99 \times 10^9) \cdot (5.00 \times 10^{-6})}{7.00^2} = 9.183 \times 10^2 \, \text{N/C}\]

Step 4: Explain Gauss's Law and Flux

Gauss's law states that the net electric flux through a closed surface is zero if there is no charge enclosed. Although the field is stronger at \(x=3.00 \, \text{m}\) than at \(x=-3.00 \, \text{m}\), the surface areas through which these fields penetrate differ. The electric field lines enter the near side more densely but exit the far side less densely over a larger area, keeping total flux balanced. The integral of \(E \, dA\) over the whole sphere, taking into account the angles and area, results in zero net flux.

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle in the study of electric fields and flux. It states that the net electric flux through any closed surface, also known as a Gaussian surface, is directly proportional to the net charge enclosed within that surface. Mathematically, it is expressed as: \[ \Phi_{E} = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] where \( \Phi_{E} \) is the electric flux, \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is the differential area vector, \( Q_{\text{enc}} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space.
Gauss's Law is particularly useful for calculating electric fields when dealing with symmetrical charge distributions, such as spherical or cylindrical distributions. In the context of this exercise, even though the electric field strength differs at different points on the sphere, the net electric flux is zero because the sphere does not enclose any charge. This concept underscores how the distribution of electric field lines over the surface maintains balance, compensating for areas with higher intensity by covering wider regions at lower intensity.

Coulomb's Law

Coulomb's Law describes the electrostatic interaction between electric charges. This law states that the magnitude of the electric force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The equation is given by: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( F \) is the magnitude of the force between the charges, \( k \) is Coulomb's constant \((8.99 \times 10^9 \text{ Nm}^2/\text{C}^2)\), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
In the provided exercise, this law helps us determine the electric field magnitude at different locations relative to a point charge. By applying Coulomb's Law, we calculate that a point at \( x = 3.00 \text{ m} \) experiences a stronger electric field compared to a point at \( x = -3.00 \text{ m} \), due to the closer proximity to the point charge. This difference illustrates how electric fields diminish rapidly with distance from a source charge.

Net Electric Flux

Net electric flux refers to the sum of electric field lines passing through a closed surface, which provides insights into the presence and distribution of electric charges. It is computed by the surface integral of the electric field over a closed surface: \[ \Phi_{E} = \oint \mathbf{E} \cdot d\mathbf{A} \] The net electric flux through a closed surface is zero if there are no charges enclosed by that surface.
In the exercise, the sphere centered at the origin does not enclose the external point charge, even though electric field lines are present across the sphere's surface. On the side closer to the charge (at \( x = 3.00 \text{ m} \)), the field lines are denser, indicating a stronger electric field. However, these lines spread out as they pass through the sphere, balancing out the flux. The total flux remains zero because the field lines that enter the sphere eventually exit, maintaining equality between entering and exiting flux.