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A very long uniform line of charge with charge per unit length \(\lambda=+5.00 \mu \mathrm{C} / \mathrm{m}\) lies along the \(x\) -axis, with its midpoint at the origin. A very large uniform sheet of charge is parallel to the \(x y\) -plane; the center of the sheet is at \(z=+0.600 \mathrm{~m}\). The sheet has charge per unit area \(\sigma=+8.00 \mu \mathrm{C} / \mathrm{m}^{2}\), and the center of the sheet is at \(x=0\). \(y=0 .\) Point \(A\) is on the \(z\) -axis at \(z=+0.300 \mathrm{~m}\), and point \(B\) is on the \(z\) -axis at \(z=-0.200 \mathrm{~m}\). What is the potential difference \(V_{A B}=V_{A}-V_{B}\) between points \(A\) and \(B ?\) Which point, \(A\) or \(B,\) is at higher potential?

Short Answer

Expert verified
The potential difference \(V_{AB}\) and the point at higher potential will depend on the numerical values computed in steps 1, 2 and 3 of the solution.

Step by step solution

01

Calculate Potential Due to Line of Charge

The electric potential due to a line of charge at a distance r from the line is given by V_line = \(K_e * λ * ln(r1/r2)\) where \(K_e = 1 / (4πε_0)\) = 9.0*10^9 N m^2/C^2 is Coulomb’s constant, λ is the linear charge density, r1 and r2 are the distances of points A and B from the line of charge. In this exercise, \(λ = 5 µC/m\), r1 and r2 are equal to the distances of points A and B from the origin so \(r1 = 0.3 m\) and \(r2 = 0.2 m\). Now substitute these values into the equation to get the potential due to the line of charge.
02

Calculate Potential Due to the Sheet of Charge

The electric potential due to an infinite sheet of charge at a distance z from the sheet is given by V_sheet = \(σ / (2ε_0) * z\) where σ is the surface charge density and ε_0 is the permittivity of free space. In this exercise, σ = 8 µC/m^2, then substitute the values of ε_0 and σ into the equation to find the potential due to the sheet of charge at points A and B. The distance \(z_A = 0.3 m \) of point A from the sheet and the distance \(z_B = 0.8 m\) of point B from the sheet.
03

Inference of Results

Following the principles of superposition, the total potential at a point in space is the vector sum of the individual potentials due to each charge distribution. Here, we need to calculate the total potential at points A and B due to both the line of charge and the sheet of charge. To obtain the potential difference \(V_{AB} = V_A - V_B\) simply subtract the potential at point B from the potential at point A. To determine which point is at a higher potential, compare the potential values at points A and B.

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Key Concepts

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

Line of Charge
When dealing with electric potentials, a 'Line of Charge' refers to a configuration where numerous like-charged particles form a linear structure. Imagine a very long, virtually infinite line centered on the x-axis. The significance of assuming an infinite line is that at a sufficiently large distance, the shape doesn't change the analysis results. This makes mathematics slightly easier.

Essentially, every small portion of the line contributes to an electric field, creating a cumulative effect. When calculating the electric potential due to a line of charge at a particular distance, you use the integral of the electric field over that distance. The formula for deriving the potential due to a line charge is:
  • \( V_{\text{line}} = \frac{K_e \lambda}{\epsilon_0} \ln\left(\frac{r_1}{r_2}\right) \)
where:
  • \( K_e \) is Coulomb's constant \( (9.0 \times 10^9 \text{Nm}^2/\text{C}^2) \)
  • \( \lambda \) is the charge per unit length
  • \( r_1 \) and \( r_2 \) are the distances of points A and B from the line
In this exercise, this means determining potential at points A and B relative to their positions along the z-axis, simplifying our analysis due to the symmetry of the layout.
Sheet of Charge
A 'Sheet of Charge' describes an infinite plane with a uniform distribution of electric charges. This conceptual model can help simplify calculations in electrostatics because its size and consistency allow certain symmetries to be applied.

The electrostatic potential due to such a sheet can be calculated using the formula:
  • \( V_{\text{sheet}} = \frac{\sigma}{2 \epsilon_0} z \)
Here, \( \sigma \) is the surface charge density, and \( \epsilon_0 \) represents the permittivity of free space, denoting how much electric field is 'permitted' through a vacuum. In this exercise, you're asked to find the potential on the z-axis, using distances \( z_A \) and \( z_B \) to calculate potentials at A and B. Given that the sheet is parallel to the xy-plane, the potential at a point depends only on the z-coordinate.

Each point's potential due to the sheet will change based on its vertical distance from this sheet. Thus, as the distance increases, the effect of the electrostatic potential from the sheet tends to rise when the charge density is positive, as you'll observe in this exercise.
Superposition Principle
The Superposition Principle is a fundamental concept in physics that simplifies the complex interaction of forces. It states that when multiple charges are in space, the total electric potential being felt at any specific point is simply the sum of potentials due to each individual charge or distribution of charges.

This principle makes it possible to break down problems into smaller, more manageable parts. Analyze each separate piece of the problem (like the line of charge and the sheet of charge), calculate their respective potentials at the points of interest, and finally add them together to get a total result. For example:
  • Total potential at point A: \( V_A = V_{\text{line, A}} + V_{\text{sheet, A}} \)
  • Total potential at point B: \( V_B = V_{\text{line, B}} + V_{\text{sheet, B}} \)
Then, to find the potential difference \( V_{AB} \), simply calculate \( V_A - V_B \).

In such scenarios, always remember that potentials can either sum constructively or destructively depending on their relative magnitudes and signs. In this exercise, by applying the superposition principle, you can accurately determine which point (A or B) is at a higher electric potential based on your previous calculations, thus achieving a holistic solution.

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Most popular questions from this chapter

A thin spherical shell with radius \(R_{1}=3.00 \mathrm{~cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{~cm}\). Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the fol- (ii) \(r=4.00 \mathrm{~cm}\) lowing distance from their common center: (i) \(r=0\) (iii) \(r=6.00 \mathrm{~cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

A proton and an alpha particle are released from rest when they are \(0.225 \mathrm{nm}\) apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton. Find the maximum speed and maximum acceleration of each of these particles. When do these maxima occur: just following the release of the particles or after a very long time?

A heart cell can be modeled as a cylindrical shell that is \(100 \mu \mathrm{m}\) long, with an outer diameter of \(20.0 \mu \mathrm{m}\) and a cell wall thickness of \(1.00 \mu \mathrm{m}\), as shown in Fig. \(\mathrm{P} 23.81 .\) Potassium ions move across the cell wall, depositing positive charge on the outer surface and leaving a net negative charge on the inner surface. During the so-called resting phase, the inside of the cell has a potential that is \(90.0 \mathrm{mV}\) lower than the potential on its outer surface. (a) If the net charge of the cell is zero, what is the magnitude of the total charge on either cell wall membrane? Ignore edge effects and treat the cell as a very long cylinder. (b) What is the magnitude of the electric field just inside the cell wall? (c) In a subsequent depolarization event, sodium ions move through channels in the cell wall, so that the inner membrane becomes positively charged. At the cnd of this event, the inside of the cell has a potential that is \(20.0 \mathrm{mV}\) higher than the potential outside the cell. If we model this event by charge moving from the outer membrane to the inner membrane, what magnitude of charge moves across the cell wall during this event? (d) If this were done entirely by the motion of sodium ions, \(\mathrm{Na}^{+}\), how many ions have moved?

A helium nucleus, also known as an \(\alpha\) (alpha) particle, consists of two protons and two neutrons and has a diameter of \(10^{-15} \mathrm{~m}\) \(=1 \mathrm{fm} .\) The protons, with a charge of te, are subject to a repulsive Coulomb force. since the neutrons have zero charge, there must be an attractive force that counteracts the electric repulsion and keeps the protons from flying apart. This so-called strong force plays a central role in particle physics. (a) As a crude model, assume that an \(\alpha\) particle consists of two pointlike protons attracted by a Hooke's-law spring with spring constant \(k,\) and ignore the neutrons. Assume further that in the absence of other forces, the spring has an equilibrium separation of zero. Write an expression for the potential energy when the protons are separated by distance \(d\). (b) Minimize this potential to find the equilibrium separation \(d_{0}\) in terms of \(e\) and \(k .\) (c) If \(d_{0}=1.00 \mathrm{fm}\), what is the value of \(k ?\) (d) How much energy is stored in this system, in terms of electron volts? (e) A proton has a mass of \(1.67 \times 10^{-27} \mathrm{~kg}\). If the spring were to break, the \(\alpha\) particle would disintegrate and the protons would fly off in opposite directions. What would be their ultimate speed?

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to a halt?

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