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Chapter 2: Problem 27

Given the electric field \(\mathbf{E}=(4 x-2 y) \mathbf{a}_{x}-(2 x+4 y) \mathbf{a}_{y}\), find \((a)\) the equation of the streamline that passes through the point \(P(2,3,-4) ;(b)\) a unit vector specifying the direction of \(\mathbf{E}\) at \(Q(3,-2,5)\).

Short Answer

Expert verified
(b) What is the unit vector specifying the direction of the electric field at point Q(3,-2,5)?

Step by step solution

01

Calculate the electric field vector for point P

Using the given electric field equation, we will find the electric field vector at point P with coordinates (2,3). Note that since the electric field has no z-component, we can disregard the z-coordinate of point P. \(\mathbf{E}_{P} = (4(2) - 2(3))\mathbf{a}_{x} - (2(2) + 4(3))\mathbf{a}_{y}\) \(\mathbf{E}_{P} = (2)\mathbf{a}_{x} - (16)\mathbf{a}_{y}\)
02

Write the streamline equation

The streamline equation is a differential equation that relates the partial derivatives of x and y with the respective components of the electric field. Since the electric field components are given by \(E_x = 4x-2y\) and \(E_y = -2x+4y\), we have: \(\frac{dy}{dx}=\frac{E_y}{E_x} = \frac{-2x + 4y}{4x - 2y}\)
03

Solve the differential equation

To solve the differential equation, rearrange the equation as follows: \((-2x + 4y)dx = (4x - 2y)dy\) Integrate both sides: \(\int (-2x + 4y)dx = \int (4x - 2y)dy\) \(-x^2 + 4xy = 2x^2 - y^2 + C\) Rearrange the equation: \(3x^2 - 4xy - y^2 = -C\)
04

Determine the constant of integration and the equation of the streamline

To find the value of C, plug in the coordinates of point P (2,3): \(3(2)^2 - 4(2)(3) - (3)^2 = -C\) \(12 - 24 - 9 = -C\) \(C = 21\) So, the equation of the streamline becomes: \(3x^2 - 4xy - y^2 = - 21\)
05

Calculate the Electric Field vector at point Q

Using the given electric field equation, we will find the electric field vector at point Q with coordinates (3,-2): \(\mathbf{E}_{Q} = (4(3) - 2(-2))\mathbf{a}_{x} - (2(3) + 4(-2))\mathbf{a}_{y}\) \(\mathbf{E}_{Q} = (16)\mathbf{a}_{x} + (2)\mathbf{a}_{y}\)
06

Find the unit vector of the electric field at point Q

To find the unit vector of the electric field at point Q, we will divide it by its magnitude: \(\lVert \mathbf{E}_{Q} \rVert = \sqrt{(16)^2 + (2)^2} = \sqrt{260}\) Then, the unit vector is: \(\hat{\mathbf{E}}_{Q} = \frac{\mathbf{E}_{Q}}{\lVert \mathbf{E}_{Q} \rVert} = \frac{16}{\sqrt{260}}\mathbf{a}_{x} + \frac{2}{\sqrt{260}}\mathbf{a}_{y}\) So, the final answers are: (a) The equation of the streamline that passes through point P(2,3,-4) is: \(3x^2 - 4xy - y^2 = - 21\) (b) The unit vector specifying the direction of \(\mathbf{E}\) at point Q(3,-2,5) is: \(\frac{16}{\sqrt{260}}\mathbf{a}_{x} + \frac{2}{\sqrt{260}}\mathbf{a}_{y}\)

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Most popular questions from this chapter

Two identical uniform sheet charges with \(\rho_{s}=100 \mathrm{nC} / \mathrm{m}^{2}\) are located in free space at \(z=\pm 2.0 \mathrm{~cm}\). What force per unit area does each sheet exert on the other?

For fields that do not vary with \(z\) in cylindrical coordinates, the equations of the streamlines are obtained by solving the differential equation \(E_{p} / E_{\phi}=\) \(d \rho /(\rho d \phi)\). Find the equation of the line passing through the point \(\left(2,30^{\circ}, 0\right)\) for the field \(\mathbf{E}=\rho \cos 2 \phi \mathbf{a}_{\rho}-\rho \sin 2 \phi \mathbf{a}_{\phi}\).

An electric dipole (discussed in detail in Section 4.7) consists of two point charges of equal and opposite magnitude \(\pm Q\) spaced by distance \(d\). With the charges along the \(z\) axis at positions \(z=\pm d / 2\) (with the positive charge at the positive \(z\) location), the electric field in spherical coordinates is given by \(\mathbf{E}(r, \theta)=\left[Q d /\left(4 \pi \epsilon_{0} r^{3}\right)\right]\left[2 \cos \theta \mathbf{a}_{r}+\sin \theta \mathbf{a}_{\theta}\right]\), where \(r>>d\). Using rectangular coordinates, determine expressions for the vector force on a point charge of magnitude \(q(a)\) at \((0,0, z) ;(b)\) at \((0, y, 0)\).

A uniform line charge of \(16 \mathrm{nC} / \mathrm{m}\) is located along the line defined by \(y=\) \(-2, z=5\). If \(\epsilon=\epsilon_{0}:\) (a) find \(\mathbf{E}\) at \(P(1,2,3) .\) (b) find \(\mathbf{E}\) at that point in the \(z=0\) plane where the direction of \(\mathbf{E}\) is given by \((1 / 3) \mathbf{a}_{y}-(2 / 3) \mathbf{a}_{z} .\)

Point charges of \(50 \mathrm{nC}\) each are located at \(A(1,0,0), B(-1,0,0), C(0,1,0)\), and \(D(0,-1,0)\) in free space. Find the total force on the charge at \(A\).
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