Question

The electric flux for gaussian surface \(\mathrm{A}\) that enclose the \(\ldots \ldots\) charged particles in free space is (given \(\left.q_{1}=-14 n c, q_{2}=78.85 \mathrm{nc}, q_{3}=-56 n c\right)\) (A) \(10^{4} \mathrm{Nm}^{2} / \mathrm{C}\) (B) \(10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (C) \(6.2 \times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (D) \(6.3 \times 10^{4} \mathrm{Nm}^{2} / \mathrm{C}\)

Short answer

The electric flux for Gaussian surface A that encloses the charged particles in free space is \(10^3 \frac{N \cdot m^2}{C}\).

Step-by-step solution

Write down the formula for electric flux for enclosed charges

The Electric Flux formula through a closed surface is given by: \(\Phi = \frac{Q_{enclosed}}{\varepsilon_0}\) Where \(\Phi\) is the electric flux, \(Q_{enclosed}\) is the charge enclosed by the Gaussian surface and \(\varepsilon_0\) is the vacuum permittivity constant, approximately equal to \(8.85 \times 10^{-12}\, \frac{C^2}{N \cdot m^2}\).

Find the total enclosed charge

The enclosed charge is the sum of the given charges, \(q_1, q_2\), and \(q_3\). Calculate the total enclosed charge as follows: \(Q_{enclosed} = q_1 + q_2 + q_3\) Given the charges: \(q_1=-14\, nC\) \(q_2=78.85\, nC\) \(q_3=-56\, nC\)

Calculate the total enclosed charge

Replace the values of \(q_1, q_2\), and \(q_3\) in the equation and solve for \(Q_{enclosed}\): \(Q_{enclosed} = (-14 + 78.85 - 56) nC\) \(Q_{enclosed} = 8.85 nC\) Notice that the result is in nano Coulombs. We need to convert this value to Coulombs. \(1\,nC = 10^{-9}C\)

Convert the total enclosed charge to Coulombs

Convert the total enclosed charge to Coulombs: \(Q_{enclosed} = 8.85 \times 10^{-9} C\)

Calculate the electric flux

Now that we have the total enclosed charge in Coulombs, substitute the value of \(Q_{enclosed}\) into the electric flux formula and calculate \(\Phi\): \(\Phi = \frac{8.85 \times 10^{-9} C}{8.85 \times 10^{-12} \frac{C^2}{N \cdot m^2}}\) \(\Phi = 10^3 \frac{N \cdot m^2}{C}\) Comparing the calculated electric flux with the given options, we find that it matches option (B). The electric flux for Gaussian surface A that encloses the charged particles in free space is \(10^3 \frac{N \cdot m^2}{C}\).

Key concepts

Gaussian Surface

A Gaussian Surface is a theoretical closed surface used in electromagnetism to help calculate electric flux. It is not a physical boundary but a mathematical construct that simplifies calculations involving electric fields. When using Gauss's Law, the Gaussian surface is chosen so that it takes advantage of symmetry in the problem area.
When deciding on the shape of a Gaussian surface, consider:

• Symmetry: Choose symmetrical shapes like spheres, cylinders, or planes to align with the symmetrical nature of the electric field.
• Enclosed Charge: The chosen surface should entirely enclose the charges to be analyzed.
• Field Lines: The surface should be aligned such that it makes calculations easier, considering how field lines enter and exit the surface.

The primary purpose of applying a Gaussian Surface is to evaluate how much electric field "flux" passes through it. It assists in deducing the relation between the inward or outward electric field lines and the charges within the surface.

Enclosed Charge

In any electrostatic problem involving a Gaussian Surface, you must calculate the Enclosed Charge. This is critical because Gauss's Law directly relates the electric flux through a closed surface to the total charge enclosed within that surface.

To find the enclosed charge, you can simply sum up all the individual charges within the Gaussian Surface:

• Identify all the charges within the Gaussian Surface.
• Add them up, respecting their signs (positive or negative) as they contribute to the total charge.

Keep in mind that the enclosed charge is often expressed in Coulombs (C), the standard unit of electric charge. However, in many problems, charges could be given in nano Coulombs (nC), so don't forget to convert these to Coulombs, such as using the conversion factor of \(1 \, \text{nC} = 10^{-9} \, \text{C}\).
Understanding the enclosed charge is crucial because it lets you accurately apply Gauss's Law to determine the resultant electric flux through the Gaussian Surface.

Vacuum Permittivity

Vacuum Permittivity, denoted as \(\varepsilon_0\), is a fundamental constant that characterizes the ability of the vacuum (or free space) to permit electric field lines. It plays a vital role in Gauss's Law and other equations dealing with electric fields in a vacuum.

Important aspects of Vacuum Permittivity include:

• Constant Value: The value of vacuum permittivity is approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\).
• Role in Gauss's Law: It links the total electric flux \( \Phi \) through a surface to the enclosed charge \( Q_{enclosed} \) via the formula: \( \Phi = \frac{Q_{enclosed}}{\varepsilon_0} \).
• Nature: As a universal constant, \(\varepsilon_0\) ensures that calculations involving electric fields and flux in vacuum can be consistently applied.

Understanding \(\varepsilon_0\) is key when calculating electric flux, especially when transitioning between charges and electric fields. It demonstrates the resistance encountered in forming an electric field in free space.