/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1569 A long string with a charge of \... [FREE SOLUTION] | 91Ó°ÊÓ

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A long string with a charge of \(\lambda\) per unit length passes through an imaginary cube of edge \(\ell\). The maximum possible flux of the electric field through the cube will be ....... (A) \(\sqrt{3}\left(\lambda \ell / \in_{0}\right)\) (B) \(\left(\lambda \ell / \in_{0}\right)\) (C) \(\sqrt{2}\left(\lambda \ell / \in_{0}\right)\) (D) \(\left[\left(6 \lambda \ell^{2}\right) / \epsilon_{0}\right]\)

Short Answer

Expert verified
The short answer is: \(\boxed{\text{(A)}\ \sqrt{3}\left(\frac{\lambda \ell}{\epsilon_{0}}\right)}\).

Step by step solution

01

Apply Gauss's law and note the electric field symmetry

Since we are dealing with electric flux through a closed surface (cube), we can use Gauss's law for electrostatics: \[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{encl}}{\epsilon_0}\] Where: - \(\vec{E}\) is the electric field vector. - \(d\vec{A}\) is the differential area vector enclosing the closed surface. - \(Q_{encl}\) is the total charge enclosed within the closed surface (cube). - \(\epsilon_0\) is the permittivity of free space (electric constant). In our case, the electric field \(\vec{E}\) is due to the charged string, and the surface is the cube of edge length â„“. As we are interested in the maximum possible electric flux, we should place the string along a space diagonal of the cube.
02

Determine the total enclosed charge

To find the total charge enclosed within the cube, we multiply the charge per unit length λ by the length of the string inside the cube (along the space diagonal): \[Q_{encl} = \lambda d\] Where \(d = \sqrt{3}\ell\) is the length of the space diagonal of the cube. Therefore, the enclosed charge is: \[Q_{encl} = \lambda \sqrt{3}\ell\]
03

Calculate the maximum electric flux

Now we can calculate the maximum electric flux (Φ) using Gauss's law: \[\Phi = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{encl}}{\epsilon_0}\] Substituting the value for \(Q_{encl}\): \[\Phi = \frac{\lambda \sqrt{3}\ell}{\epsilon_0}\] Comparing this result with the given options, we find that the maximum possible electric flux through the cube is: \(\boxed{\text{(A)}\ \sqrt{3}\left(\frac{\lambda \ell}{\epsilon_{0}}\right)}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Flux
Electric flux is an essential concept when understanding how an electric field interacts with a surface. It measures the number of electric field lines passing through any given surface. Think of it like wind passing through a net; each line of wind represents the electric field lines.
In mathematical terms, electric flux \( \Phi \) is calculated by:
  • Integrating the electric field \( \vec{E} \) over a given surface with area element \( d\vec{A} \), resulting in \( \Phi = \oint \vec{E} \cdot d\vec{A} \).
This dot product signifies the component of the electric field that is passing through the surface. Therefore, if the field lines are perpendicular to the surface, the flux is maximum, and if parallel, the flux is zero.
Gauss's Law is pivotal here, relating electric flux to the charge enclosed by a surface. According to Gauss's Law, flux \( \Phi \) through a closed surface is: \ \[\Phi = \frac{Q_{encl}}{\epsilon_0} \]\,where \( Q_{encl} \) is the enclosed charge, a crucial part of determining electric flux in various scenarios.
Permittivity of Free Space
The permittivity of free space, commonly denoted by \( \epsilon_0 \), is a fundamental physical constant that plays a crucial role in electrostatics. It quantifies how much electric field flux is generated per unit charge in a vacuum. Imagine it as a measure of how easily electric fields "pass through" free space.
Its value is approximately \( 8.854 \times 10^{-12} \ \text{F/m} \) (farads per meter). This small number indicates that free space does not allow electric fields to pass as easily compared to materials with higher permittivity. In Gauss's Law, \( \epsilon_0 \) appears in the denominator of the equation \( \Phi = \frac{Q_{encl}}{\epsilon_0} \). Hence, it determines the size of the electric flux based on the enclosed charge.
  • Materials with higher permittivity can store more charge at a lower voltage, impacting how electric fields interact with different materials.
  • In a vacuum, which has an \( \epsilon_0 \), electric fields interact in a relatively "unhindered" manner compared to other materials.
Charge Distribution on Surfaces
Charge distribution on surfaces influences how electric fields interact and how flux is calculated. Imagine that charges are spread out uniformly along a line, surface, or volume, affecting how the electric field emanates from these charges.
For linear charge distribution, such as a string with charge distributed along its length as \( \lambda \) (charge per unit length), the electric field lines spread out in radial patterns from the string. This specific distribution is essential when using Gauss's Law to determine flux through a surface.
  • In this problem, the string's charge affects the electric flux passing through a cube based on how it aligns with the cube's geometry.
  • When the string is aligned along the space diagonal of the cube, it maximizes the enclosed charge using the formula \( Q_{encl} = \lambda \sqrt{3} \ell \).
This alignment creates the maximum electric field interaction possible with the surface of the cube, directly influencing the amount of electric flux as calculated through Gauss’s Law. Understanding this helps in predicting and analyzing the behavior of electric fields in various configurations.

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Most popular questions from this chapter

The circular plates \(\mathrm{A}\) and \(\mathrm{B}\) of a parallel plate air capacitor have a diameter of \(0.1 \mathrm{~m}\) and are \(2 \times 10^{-3} \mathrm{~m}\) apart. The plates \(\mathrm{C}\) and \(\mathrm{D}\) of a similar capacitor have a diameter of \(0.1 \mathrm{~m}\) and are \(3 \times 10^{-3} \mathrm{~m}\) apart. Plate \(\mathrm{A}\) is earthed. Plates \(\mathrm{B}\) and \(\mathrm{D}\) are connected together. Plate \(\mathrm{C}\) is connected to the positive pole of a \(120 \mathrm{~V}\) battery whose negative is earthed, The energy stored in the system is (A) \(0.1224 \mu \mathrm{J}\) (B) \(0.2224 \mu \mathrm{J}\) (C) \(0.4224 \mu \mathrm{J}\) (D) \(0.3224 \mu \mathrm{J}\)

At what angle \(\theta\) a point \(P\) must be located from dipole axis so that the electric field intensity at the point is perpendicular to the dipole axis? (A) \(\tan ^{-1}(1 / \sqrt{2})\) (B) \(\tan ^{-1}(1 / 2)\) (C) \(\tan ^{-1}(2)\) (C) \(\tan ^{-1}(\sqrt{2})\)

Let \(\mathrm{P}(\mathrm{r})\left[\mathrm{Q} /\left(\pi \mathrm{R}^{4}\right)\right] \mathrm{r}\) be the charge density distribution for a solid sphere of radius \(\mathrm{R}\) and total charge \(\mathrm{Q}\). For a point ' \(\mathrm{P}\) ' inside the sphere at distance \(\mathrm{r}_{1}\) from the centre of the sphere the magnitude of electric field is (A) \(\left[\mathrm{Q} /\left(4 \pi \epsilon_{0} \mathrm{r}_{1}^{2}\right)\right]\) (B) \(\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(4 \pi \in{ }_{0} \mathrm{R}^{4}\right)\right]\) (C) \(\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(3 \pi \epsilon_{0} \mathrm{R}^{4}\right)\right]\)

The capacities of three capacitors are in the ratio \(1: 2: 3\). Their equivalent capacity when connected in parallel is \((60 / 11) \mathrm{F}\) more then that when they are connected in series. The individual capacitors are of capacities in \(\mu \mathrm{F}\) (A) \(4,6,7\) (B) \(1,2,3\) (C) \(1,3,6\) (D) \(2,3,4\)

Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases \(4.5\) times in comparison with the initial value. The ratio of the initial charges of the balls is ....... (A) \(4: 1\) (B) \(6: 1\) (C) \(3: 1\) (D) \(2: 1\)

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