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Chapter 23: Problem 45

In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x-y-\) , and \(z\) -components of the electric field. (b) At which points is the electric field equal to zero?

Short Answer

Expert verified
The electric field is zero at \(x = -\frac{C}{A}\), \(y = -2\frac{BC}{A^2}\), for any \(z\).

Step by step solution

01

Understanding Electric Field Components

The electric field components are the negative partial derivatives of the electric potential with respect to each coordinate. This means we have to differentiate the given electric potential function with respect to each of the variables: \(x\), \(y\), and \(z\).
02

Calculating the Electric Field Component \(E_x\)

Compute the partial derivative of the potential \(V(x,y,z) = Axy - Bx^2 + Cy\) with respect to \(x\).\[E_x = -\frac{\partial V}{\partial x} = -(Ay - 2Bx).\]
03

Calculating the Electric Field Component \(E_y\)

Compute the partial derivative of the potential \(V(x,y,z)\) with respect to \(y\).\[E_y = -\frac{\partial V}{\partial y} = -(Ax + C).\]
04

Calculating the Electric Field Component \(E_z\)

Compute the partial derivative of the potential \(V(x,y,z)\) with respect to \(z\). Since \(z\) is not in the expression for \(V\), its derivative is zero.\[E_z = -\frac{\partial V}{\partial z} = 0.\]
05

Finding Points Where Electric Field is Zero

Set the expressions for \(E_x\), \(E_y\), and \(E_z\) equal to zero and solve for \(x\), \(y\), and \(z\). For \(E_x = 0: Ay - 2Bx = 0 \rightarrow y = \frac{2B}{A}x.\)For \(E_y = 0: Ax + C = 0 \rightarrow x = -\frac{C}{A}.\)For \(E_z = 0:\) any \(z\) since it's already zero-independent of \(z\). Combine the above to find that the point where the electric field is zero is where \(x = -\frac{C}{A}\) and \(y = -2\frac{BC}{A^2}\) for any \(z\).

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Key Concepts

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

Electric Potential
Electric potential is a fundamental concept in electrostatics, representing the potential energy per unit charge at a given point in a field. It is a scalar quantity, meaning it only has magnitude and no direction.
The potential is usually denoted by \(V\) and is related to the electric field \(\mathbf{E}\) through negative derivatives.
In this exercise, the potential is given as a function \(V(x, y, z) = Axy - Bx^2 + Cy\). Here, \(A, B,\) and \(C\) are positive constants that scale the influence of \(x\) and \(y\) within the potential field.
  • Electric potential allows us to calculate electric fields, which describe how charges interact at different points in space.
  • Potential affects how work is done by or against electric forces.
  • Uniform potentials translate to constant electric fields, whereas varying potentials indicate changes in field strength and direction.
Understanding how electric potential functions lay the groundwork for predicting particle behavior in a field.
Partial Derivatives
Partial derivatives measure how a multivariable function changes as one variable changes while keeping the others constant.
This concept is instrumental in finding electric fields from electric potential functions.
To derive the components of the electric field from \(V(x, y, z)\), we compute the partial derivatives with respect to \(x, y,\) and \(z\).
  • The partial derivative with respect to \(x\) gives us the rate of change of the potential that is influenced by changes in \(x\).
  • Likewise, differentiating with \(y\) tells us how the potential changes with respect to \(y\).
  • For \(z\), if \(z\) does not appear in the potential function, its derivative will be zero, indicating no change in potential with \(z\).
The calculated derivatives directly yield the electric field components through the relation \(E_i = -\frac{\partial V}{\partial i}\), where \(i\) represents each direction \(x, y, z\).
Vector Components
Electric fields are vector quantities, meaning they possess both magnitude and direction.
The field can be described by its components along the \(x, y,\) and \(z\) axes, corresponding to how much the field acts in each direction.
In our example, the electric field \(\mathbf{E}\) is broken down into components \(E_x, E_y,\) and \(E_z\), which are derived from the electric potential through partial derivatives.
  • \(E_x = -(Ay - 2Bx)\) represents the field's impact in the horizontal direction.
  • \(E_y = -(Ax + C)\) captures the field's influence vertically.
  • \(E_z = 0\) shows no component in the \(z\)-direction because the potential does not vary with \(z\).
Understanding vector components is crucial to picture how the field exerts forces on charged particles in space.
Electrostatics
Electrostatics is the study of electric charges at rest and the electric fields and potentials generated by these charges.
This branch of physics helps us to understand how charges interact in empty space or in the presence of conductors and dielectric materials.
Using electrostatics, we can predict how charges will behave when positioned in a field based on the electric potential function.
  • The electric field is a vector field that shows the force exerted per unit charge at any point in space.
  • Electric potential helps us determine the work done during the movement of charges in the field.
  • With these principles, we can find conditions such as equilibrium points where the field exerts no force, as determined by setting field components to zero.
In this exercise, electrostatics was harnessed to compute potential-derived field components and find zero-field conditions.

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Most popular questions from this chapter

A small particle has charge \(-5.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-4} \mathrm{kg} .\) It moves from point \(A,\) where the electric potential is \(V_{A}=+200 \mathrm{V},\) to point \(B,\) where the electric potential is \(V_{B}=+800 \mathrm{V} .\) The electric force is the only force acting on the particle. The particle has speed 5.00 \(\mathrm{m} / \mathrm{s}\) at point \(A .\) What is its speed at point \(B ?\) Is it moving faster or slower at \(B\) than at \(A ?\) Explain.

Two point charges \(q_{1}=+2.40 \mathrm{nC} \quad\) and \(\quad q_{2}=\) \(-6.50 \mathrm{nC}\) are 0.100 \(\mathrm{m}\) apart. Point \(A\) is midway between them; point \(B\) is 0.080 \(\mathrm{m}\) from \(q_{1}\) and 0.060 \(\mathrm{m}\) from \(q_{2}\) (Fig. E23.19). Take the electric potential to be zero at infinity. Find (a) the potential at point \(A\); (b) the potential at point \(B ;\) (c) the work done by the electric field on a charge of 2.50 \(\mathrm{nC}\) that travels from point \(B\) to point \(A\) .

A very long cylinder of radius 2.00 \(\mathrm{cm}\) carries a uniform charge density of 1.50 \(\mathrm{nC} / \mathrm{m}\) (a) Describe the shape of the equipotential surfaces for this cylinder. (b) Taking the reference level for the zero of potential to be the surface of the cylinder, find the radius of equipotential surfaces having potentials of \(10.0 \mathrm{V}, 20.0 \mathrm{V},\) and 30.0 \(\mathrm{V}\) . (c) Are the equipotential surfaces equally spaced? If not, do they get closer together or farther apart as \(r\) increases?

Self-Energy of a Sphere of Charge. A solid sphere of radius \(R\) contains a total charge \(Q\) distributed uniformly throughout its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the "self-energy" of the charge distribution. (Hint: After you have assembled a charge \(q\) in a sphere of radius \(r,\) how much energy would it take to add a spherical shell of thickness dr having charge \(d q ?\) Then integrate to get the total energy.)

A small sphere with mass \(5.00 \times 10^{-7} \mathrm{kg}\) and charge \(+3.00 \mu C\) is released from rest a distance of 0.400 \(\mathrm{m}\) above a large horizontal insulating sheet of charge that has uniform surface charge density \(\sigma=+8.00 \mathrm{pC} / \mathrm{m}^{2} .\) Using energy methods, calculate the speed of the sphere when it is 0.100 \(\mathrm{m}\) above the sheet of charge?
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