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In a certain region of space the electric potential is given by \(V=+A x^{2} y-B x y^{2},\) where \(A=5.00 \mathrm{~V} / \mathrm{m}^{3}\) and \(B=8.00 \mathrm{~V} / \mathrm{m}^{3}\) Calculate the magnitude and direction of the electric field at the point in the region that has coordinates \(x=2.00 \mathrm{~m}, y=0.400 \mathrm{~m},\) and \(z=0\)

Short Answer

Expert verified
The magnitude and direction of the electric field at the given point can be calculated using the above steps. Remember to calculate the X and Y components of the electric field then find their magnitude and direction.

Step by step solution

01

Understand the Formula

The electric field \( E \) is related to the electric potential \( V \) by the formula: \( E = - \nabla V \) where \( \nabla V \) is gradient of the potential function.
02

Calculate the X and Y Components of Electric Field

The X and Y components of the electric field are given by: \( E_x = -\frac{\partial V}{\partial x} \)\( E_y = -\frac{\partial V}{\partial y} \)Substitute the expression for \( V \) into these equations and differentiate with respect to \( x \) and \( y \) respectively. Using the given potential function:\( E_x = - [2Ax-By^{2}] = -[2*5.00*(2.00) - 8.00*(0.400)^{2}] \) and\( E_y = - [Ax^{2} - 2Bxy] = -[5.00*(2.00)^{2} - 8.00*(2.00)*(0.400)] \)Calculate to get the X and Y components of the electric field.
03

Calculate the Magnitude and Direction of Electric Field

The magnitude of the electric field can be calculated using: \( E = \sqrt{E_x^2 + E_y^2 + E_z^2} \)In this case, because the system is two-dimensional, the Z-component of electric field can be taken as zero.Use the calculated values of \( E_x \) ,\( E_y \) and \( E_z = 0 \) to find \( E \)The direction of electric field (Theta) in degrees can be found as:Theta = arctan(abs(E_y)/abs(E_x))Theta should be evaluated with respect to the positive x-axis

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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 scalar quantity that represents the electric potential energy per unit charge at a point in a field. It simplifies the understanding of electric fields by allowing us to consider electric potential energy changes rather than forces directly. In the given problem, the electric potential is expressed as a function of spatial coordinates: \[ V = +A x^{2}y - B x y^{2} \]Here, the constants \( A = 5.00 \, \text{V}/\text{m}^{3} \) and \( B = 8.00 \, \text{V}/\text{m}^{3} \) are used to describe the variation of potential with respect to \(x\) and \(y\).
  • The electric potential is higher in regions where the configuration of \(x\) and \(y\) makes the expression larger.
  • An electric field is derived from the change in this potential.
Considering electric potential as a starting point helps establish the subsequent changes in space, which give rise to the electric field.
Electric Field Components
The electric field in any region can be broken down into components, usually expressed as \( E_x, E_y, \text{and } E_z \), corresponding to three spatial dimensions. These components tell us how much force a charge would experience at that point, in each respective direction.In the exercise:
  • The \(E_x\) component is derived by differentiating the potential \( V \) with respect to \(x\), resulting in \( E_x = - \frac{\partial V}{\partial x} \).
  • Similarly, \(E_y\) is found by differentiating \( V \) with respect to \(y\), as \( E_y = - \frac{\partial V}{\partial y} \).
This differentiation reflects the change in potential as you move in the \(x\) and \(y\) directions. The idea is that electric fields point towards areas of decreasing potential. By calculating these components separately, it's easier to understand their individual effects and combine them to find the full field at any point.
Gradient
The gradient is a key mathematical concept in understanding changes in fields, especially scalar fields like electric potential. It essentially describes the slope or incline of the potential at any given point. In mathematical terms:\[ abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \]For electric fields, the gradient at a point gives the direction and magnitude of the maximum rate of change of the potential. In simple terms, it points from areas of high potential to low potential – this is why the field vector \( \mathbf{E} \) is typically the negative of the potential’s gradient:
  • The negative sign means the field points in the direction of greatest potential decrease.
  • This can be visualized as arrows pointing downhill on a map of potential.
In the exercise, finding \( abla V \) helps determine both the direction and strength of the electric field at a specific point.
Potential Function
A potential function is a mathematical expression that helps in evaluating the potential energy of a system in varying spatial dimensions. It provides insight into how electric charges interact spatially. The given potential function:\[ V = +A x^{2}y - B x y^{2} \]allows for precise calculation of potential values at any \((x, y, z)\) point, which in turn leads to understanding the nature of the electric field at those points.
  • This function clearly shows dependency on the \(x\) and \(y\) variables, indicating that the electric potential, and hence the field, will vary with changes in these coordinates.
  • By manipulating this function through differentiation, we can unlock details like the specific changes to electric potential that result in electric fields through the components \(E_x\) and \(E_y\).
These computations are not only theoretical exercises but also serve practical applications, such as determining field intensity accessible to measure in experiments. Understanding the potential function allows a deeper grasp of how electric fields behave spatially and are fundamental for various electrostatic applications.

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Most popular questions from this chapter

A very long insulating cylinder of charge of radius \(2.50 \mathrm{~cm}\) carries a uniform linear density of \(15.0 \mathrm{nC} / \mathrm{m}\). If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads \(175 \mathrm{~V} ?\)

The maximum voltage at the center of a typical tandem electrostatic accelerator is \(6.0 \mathrm{MV}\). If the distance from one end of the acceleration tube to the midpoint is \(12 \mathrm{~m}\), what is the magnitude of the average electric field in the tube under these conditions? (a) \(41,000 \mathrm{~V} / \mathrm{m}\) (b) \(250.000 \mathrm{~V} / \mathrm{m} ;\) (c) \(500.000 \mathrm{~V} / \mathrm{m} ;\) (d) \(6.000 .000 \mathrm{~V} / \mathrm{m} .\)

The electric potential \(V\) in a region of space is given by $$ V(x, y, z)=A\left(x^{2}-3 y^{2}+z^{2}\right) $$ where \(A\) is a constant. (a) Derive an expression for the electric field \(\vec{E}\) at any point in this region. (b) The work done by the field when a \(1.50 \mu \mathrm{C}\) test charge moves from the point \((x, y, z)=(0,0,0.250 \mathrm{~m})\) to the origin is measured to be \(6.00 \times 10^{-5}\) J. Determine \(A\). (c) Determine the electric field at the point \((0,0,0.250 \mathrm{~m})\). (d) Show that in every plane parallel to the \(x z\) -plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to \(V=1280 \mathrm{~V}\) and \(y=2.00 \mathrm{~m} ?\)

A very long insulating cylindrical shell of radius \(6.00 \mathrm{~cm}\) carries charge of linear density \(8.50 \mu \mathrm{C} / \mathrm{m}\) spread uniformly over its outer surface. What would a voltmeter read if it were connected between (a) the surface of the cylinder and a point \(4.00 \mathrm{~cm}\) above the surface, and (b) the surface and a point \(1.00 \mathrm{~cm}\) from the central axis of the cylinder?

A solid conducting sphere has net positive charge and radius \(R=0.400 \mathrm{~m}\). At a point \(1.20 \mathrm{~m}\) from the center of the sphere, the electric potential due to the charge on the sphere is \(24.0 \mathrm{~V}\). Assume that \(V=0\) at an infinite distance from the sphere. What is the electric potential at the center of the sphere?

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