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An infinite plane of charge has a uniform charge distribution of \(+4.00 \mathrm{nC} / \mathrm{m}^{2}\) and is located in the \(y z\) -plane at \(x=0 . A+11.0 \mathrm{nC}\) fixed point charge is located at \(x=+2.00 \mathrm{~m}\) a) Find the electric potential \(V(x)\) on the \(x\) -axis from \(0

Short Answer

Expert verified
Answer: To find the minimum electric potential location, follow Step 4 by differentiating the total electric potential with respect to x and setting the result to zero. Solve for x to find the position(s) on the x-axis where the electric potential is minimum. To determine where a positive point charge can be placed without moving, follow Step 5 by computing the total electric field by finding the derivative of the total electric potential and setting it equal to zero. Find the position(s) on the x-axis where a positive point charge can be placed without moving.

Step by step solution

01

Find the electric potential due to the infinite plane of charge for 0 < x < 2.00 m

Recall that the electric potential $$V(x)$$ due to an infinite plane of charge is given by the formula: $$ V(x) = \frac{\sigma}{2\epsilon_0} x $$ where $$\sigma$$ is the charge density, $$\epsilon_0$$ is the vacuum permittivity (approx. $$8.854 \times 10^{-12} \frac{\mathrm{C}^2}{\mathrm{N} \cdot \mathrm{m}^2}$$), and $$x$$ is the distance from the plane. The given charge density is $$\sigma = +4.00 \mathrm{nC}/\mathrm{m}^2 = 4.00 \times 10^{-9} \mathrm{C}/\mathrm{m}^2$$. Now plug in the values into the formula: $$ V(x) = \frac{4.00 \times 10^{-9} \mathrm{C/m}^2}{2(8.854 \times 10^{-12}\mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2)}x $$
02

Find the electric potential due to the +11.0 nC point charge for 0 < x < 2.00 m

The electric potential $$V(x)$$ due to a point charge is given by the formula: $$ V(x) = \frac{kQ}{x} $$ where $$k$$ is Coulomb's constant (approx. $$8.987 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2$$), $$Q$$ is the value of the point charge, and $$x$$ is the distance from the charge. The given point charge is $$Q = +11.0 \mathrm{nC} = 11.0 \times 10^{-9} \mathrm{C}$$. Now plug in the values into the formula: $$ V(x) = \frac{8.987 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2 \cdot 11.0 \times 10^{-9} \mathrm{C}}{x} $$
03

Find the total electric potential for 0 < x < 2.00 m

The total electric potential $$V_\text{tot}(x)$$ at a point on the x-axis is the sum of the electric potentials due to the infinite plane and the point charge: $$ V_\text{tot}(x) = \frac{4.00 \times 10^{-9} \mathrm{C/m}^2}{2(8.854 \times 10^{-12}\mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2)}x + \frac{8.987 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2 \cdot 11.0 \times 10^{-9} \mathrm{C}}{x} $$
04

Find the minimum electric potential

To find the minimum electric potential, differentiate $$V_\text{tot}(x)$$ with respect to $$x$$ and set the result to zero: $$ \frac{dV_\text{tot}(x)}{dx} = 0 $$ Solving for $$x$$ will give us the position(s) on the $$x$$-axis where the electric potential is minimum.
05

Determine where a positive point charge could be placed without moving

A positive point charge can be placed without moving at a point where the electric field is zero. Recall that the electric field $$E(x)$$ is related to the electric potential by the derivative: $$ E(x) = -\frac{dV(x)}{dx} $$ Compute the total electric field by finding the derivative of the total electric potential and set it equal to zero to find the position(s) on the $$x$$-axis where a positive point charge could be placed without moving. By following these steps, you will find the electric potential on the $$x$$-axis between 0 and 2.00 m, the location(s) of the minimum electric potential, and the position(s) where a positive point charge can be placed without moving.

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Key Concepts

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

Uniform Charge Distribution
Imagine a vast, flat surface where every square meter holds precisely the same amount of charge. This is what we call a uniform charge distribution. The elegance of this concept lies in its simplicity; it allows us to predict how an electric potential or field behaves around any place in its vicinity. The uniformity in distribution means that no matter how close or far you are from the surface, the amount of charge per unit area remains constant. This consistency is vital for the accurate calculation of potentials and fields, serving as a cornerstone for more complex scenarios in electrostatics.

When dealing with an infinite plane, the notion of 'infinity' simplifies the mathematics. Why? Because at any finite distance, the edges of the plane are still so far away, the plane appears effectively infinite, maintaining a constant charge per unit area regardless of the observer's location relative to the plane.
Coulomb's Law
The heart of electrostatics beats to the rhythm of Coulomb's law, a principle defining the force between two point charges. It was named after Charles-Augustin de Coulomb, who first described it in the 1780s. The law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula reflecting Coulomb's law is both elegant and provides a clear quantification of this force: \( F = k \frac{q_1 q_2}{r^2} \) where \( F \) is the force between the charges, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant. This foundational law influences how charges interact across the vacuum, as well as within different mediums.
Electric Field
Imagine you have a map showing how a charged particle would move through space. This 'map' is what physicists refer to as the electric field. It showcases the vector force that a positive unit charge would experience at various points. The electric field is a way to represent how the presence of charge can influence the space around it, imparting a force onto other charges within the field.

Mathematically, the electric field due to a single point charge is expressed as: \( E = k \frac{q}{r^2} \) where \( E \) is the magnitude of the electric field, \( q \) is the charge creating the field, \( r \) is the distance from the charge to the point in space, and again, \( k \) is Coulomb's constant. The direction of the electric field is always radial from the point charge and outward for positive charges or inward for negative charges.
Potential Due to Point Charge
The concept of potential due to a point charge introduces us to another aspect of electrostatic fields—the potential energy per unit charge at a particular point in space. This potential is a scalar quantity, meaning it doesn't have a direction but rather represents the magnitude of potential energy a positive test charge would have at that location.

Defining Electric Potential

It is given by the formula: \( V = k \frac{q}{r} \) where \( V \) represents the electric potential, \( q \) is the point charge, \( r \) is the distance from the point charge, and \( k \) is Coulomb’s constant. Intuitively, you can think of electric potential as the electric 'pressure' at a point in space due to the charge's presence. The higher the charge or closer to it you are, the more 'pressure' or potential you have.
Potential Due to Infinite Plane
An infinite plane with a uniform charge distribution is a bit of a thought experiment, but it's incredibly handy for understanding electric potential in more manageable terms. Unlike a point charge, the potential due to an infinite plane doesn't decrease with distance. Instead, it increases linearly with the distance from the plane, because the electric field is constant.

For an infinite plane of charge, the potential is calculated using the formula: \( V = \frac{\sigma}{2\varepsilon_0} x \) where \( V \) is the potential at distance \( x \) from the plane, \( \sigma \) is the surface charge density, and \( \varepsilon_0 \) is the vacuum permittivity. This linear relationship is unique to the idealized concept of an infinite plane and illustrates how such uniformity creates simplicity in otherwise complex electrostatic situations.

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Most popular questions from this chapter

The electron beam emitted by an electron gun is controlled (steered) with two sets of parallel conducting plates: a horizontal set to control the vertical motion of the beam, and a vertical set to control the horizontal motion of the beam. The beam is emitted with an initial velocity of \(2.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\). The width of the plates is \(d=5.00 \mathrm{~cm}\), the separation between the plates is \(D=4.00 \mathrm{~cm},\) and the distance between the edge of the plates and a target screen is \(L=40.0 \mathrm{~cm}\) In the absence of any applied voltage, the electron beam hits the origin of the \(x y\) -coordinate system on the observation screen. What voltages need to be applied to the two sets of plates for the electron beam to hit a target placed on the observation screen at coordinates \((x, y)=(0 \mathrm{~cm}, 8.00 \mathrm{~cm}) ?\)

Using Gauss's Law and the relation between electric potential and electric field, show that the potential outside a uniformly charged sphere is identical to the potential of a point charge placed at the center of the sphere and equal to the total charge of the sphere. What is the potential at the surface of the sphere? How does the potential change if the charge distribution is not uniform but has spherical (radial) symmetry?

Two point charges are located at two corners of a rectangle, as shown in the figure. a) What is the electric potential at point \(A ?\) b) What is the potential difference between points \(A\) and \(B ?\)

High-voltage power lines are used to transport electricity cross country. These wires are favored resting places for birds. Why don't the birds die when they touch the wires?

A point charge \(Q\) is placed a distance \(R\) from the center of a conducting sphere of radius \(a,\) with \(R>a\) (the point charge is outside the sphere). The sphere is grounded, that is, connected to a distant, unlimited source and/or sink of charge at zero potential. (Neither the distant ground nor the connection directly affects the electric field in the vicinity of the charge and sphere.) As a result, the sphere acquires a charge opposite in sign to \(Q\), and the point charge experiences an attractive force toward the sphere. a) Remarkably, the electric field outside the sphere is the same as would be produced by the point charge \(Q\) plus an imaginary mirror-image point charge \(q\), with magnitude and location that make the set of points corresponding to the surface of the sphere an equipotential of potential zero. That is, the imaginary point charge produces the same field contribution outside the sphere as the actual surface charge on the sphere. Calculate the value and location of \(q\). (Hint: By symmetry, \(q\) must lie somewhere on the axis that passes through the center of the sphere and the location of \(Q .)\) b) Calculate the force exerted on point charge \(Q\) and directed toward the sphere, in terms of the original quantities \(Q, R,\) and \(a\) c) Determine the actual nonuniform surface charge distribution on the conducting sphere.

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