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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

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by \(45.0 \mathrm{~mm}\), and the potential difference between them is \(360 \mathrm{~V}\). (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge \(+2.40 \mathrm{nC} ?\) (c) Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower. (d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

The Millikan Oil-Drop Experiment. The charge of an electron was first measured by the American physicist Robert Millikan during \(1909-1913 .\) In his experiment, oil was sprayed in very fine drops (about \(10^{-4} \mathrm{~mm}\) in diameter) into the space between two parallel horizontal plates separated by a distance \(d\). A potential difference \(V_{A B}\) was maintained between the plates, causing a downward electric field between them. Some of the oil drops acquired a negative charge because of frictional effects or because of ionization of the surrounding air by x rays or radioactivity. The drops were observed through a microscope. (a) Show that an oil drop of radius \(r\) at rest between the plates remained at rest if the magnitude of its charge was \(q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}\) where \(\rho\) is oil's density. (Ignore the buoyant force of the air.) By adjusting \(V_{A B}\) to keep a given drop at rest, Millikan determined the charge on that drop. provided its radius \(r\) was known. (b) Millikan's oil drops were much too small to measure their radii directly. Instead, Millikan determincd \(r\) by cutting off the electric field and measuring the terminal speed \(u_{\mathrm{i}}\) of the drop as it fell. (We discussed terminal speed in Section \(\left.5.3 .\right)\) The viscous force \(F\) on a sphere of radius \(r\) moving at speed \(v\) through a fluid with viscosity \(\eta\) is given by Stokes's law: \(F=6 \pi \eta r v .\) When a drop fell at \(v_{1}\), the viscous force just balanced the drop's weight \(w=m g\). Show that the magnitude of the charge on the drop was \(q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{1}^{3}}{2 \rho g}}\) (c) You repeat the Millikan oil-drop experiment. Four of your measured values of \(V_{A B}\) and \(v_{1}\) are listed in the table: $$ \begin{array}{lcccc} \text { Drop } & 1 & 2 & 3 & 4 \\ \hline V_{A B}(\mathrm{~V}) & 9.16 & 4.57 & 12.32 & 6.28 \\ v_{1}\left(10^{-5} \mathrm{~m} / \mathrm{s}\right) & 2.54 & 0.767 & 4.39 & 1.52 \end{array} $$ In your apparatus, the separation \(d\) between the horizontal plates is \(1.00 \mathrm{~mm}\). The density of the oil you use is \(824 \mathrm{~kg} / \mathrm{m}^{3}\). For the viscosity \(\eta\) of air, use the value \(1.81 \times 10^{-5} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\). Assume that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\). Calculate the charge \(q\) of each drop. (d) If electric charge is quantized (that is, exists in multiples of the magnitude of the charge of an electron), then the charge on each drop is -ne, where \(n\) is the number of excess electrons on each drop. (All four drops in your table have negative charge.) Drop 2 has the smallest magnitude of charge observed in the experiment, for all 300 drops on which measurements were made, so assume that its charge is due to an excess charge of one electron. Determine the number of excess electrons \(n\) for each of the other three drops. (e) Use \(q=-n e\) to calculate \(e\) from the data for each of the four drops, and average these four values to get your best experimental value of \(e .\)

A solid conducting sphere of radius \(5.00 \mathrm{~cm}\) carries a net charge. To find the value of the charge, you measure the potential difference \(V_{A B}=V_{A}-V_{B}\) between point \(A,\) which is \(8.00 \mathrm{~cm}\) from the center of the sphere, and point \(B\), which is a distance \(r\) from the center of the sphere. You repeat these measurements for several values of \(r>8.00 \mathrm{~cm} .\) When you plot your data as \(V_{A B}\) versus \(1 / r,\) the values lie close to a straight line with slope \(-18.0 \mathrm{~V} \cdot \mathrm{m}\). What does your data give for the net charge on the sphere? Is the net charge positive or negative?

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?

A particle with charge \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{E}\) directed to the left. The charge is released from rest and moves to the left; after it has moved \(6.00 \mathrm{~cm},\) its kinetic energy is \(+2.20 \times 10^{-6} \mathrm{~J}\). What are (a) the work done by the electric force, (b) the potential of the starting point with respect to the end point, and (c) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) ?

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