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Chapter 1: Problem 25

Given point \(P\left(r=0.8, \theta=30^{\circ}, \phi=45^{\circ}\right)\) and \(\mathbf{E}=1 / r^{2}\left[\cos \phi \mathbf{a}_{r}+\right.\) \(\left.(\sin \phi / \sin \theta) \mathbf{a}_{\phi}\right]\), find \((a) \mathbf{E}\) at \(P ;(b)|\mathbf{E}|\) at \(P ;(c)\) a unit vector in the direction of \(\mathbf{E}\) at \(P\)

Short Answer

Expert verified
Question: Find the vector \(\mathbf{E}\) at point \(P\), its magnitude, and a unit vector in the direction of \(\mathbf{E}\) at point \(P\), given \(\mathbf{E}=1 / r^{2}\left[\cos \phi \mathbf{a}_{r}+(\sin \phi / \sin \theta) \mathbf{a}_{\phi}\right]\) and \(P(r=0.8, \theta=30^{\circ}, \phi=45^{\circ})\). Answer: The vector \(\mathbf{E}\) at point \(P\) is given by \(\mathbf{E}_{P} = \frac{5\sqrt{2}}{8}\mathbf{a}_r + \frac{5\sqrt{2}}{4}\mathbf{a}_{\phi}\). Its magnitude is \(|\mathbf{E}_{P}| = 2.5\sqrt{2}\). The unit vector in the direction of \(\mathbf{E}\) at point \(P\) is \(\mathbf{\hat{E}}_P = \frac{1}{5} \mathbf{a}_r + \frac{2}{5}\mathbf{a}_{\phi}\).

Step by step solution

01

Find the vector \(\mathbf{E}\) at point \(P\)

Given \(\mathbf{E}=1 / r^{2}\left[\cos \phi \mathbf{a}_{r}+\left.(\sin \phi / \sin \theta) \mathbf{a}_{\phi}\right].\) At point \(P(r=0.8, \theta=30^{\circ}, \phi=45^{\circ})\), we can substitute the given values into \(\mathbf{E}\) and get: \(\mathbf{E}_{P} = \frac{1}{(0.8)^2} [\cos(45^{\circ})\mathbf{a}_r + \frac{\sin(45^{\circ})}{\sin(30^{\circ})}\mathbf{a}_{\phi}]\) Now, simplify the expression: \(\mathbf{E}_{P} = \frac{1}{0.64} [\frac{\sqrt{2}}{2}\mathbf{a}_r + \sqrt{2}\mathbf{a}_{\phi}]\) \(\mathbf{E}_{P} = \frac{5\sqrt{2}}{8}\mathbf{a}_r + \frac{5\sqrt{2}}{4}\mathbf{a}_{\phi}\)
02

Find the magnitude of \(\mathbf{E}\) at point \(P\)

To find the magnitude, we first need to find the components of \(\mathbf{E}\) in rectangular coordinates. We have: \(\mathbf{E}_{P} = \frac{5\sqrt{2}}{8}\mathbf{a}_r + \frac{5\sqrt{2}}{4}\mathbf{a}_{\phi}\) Now, calculate the magnitude as follows: \(|\mathbf{E}_{P}| = \sqrt{\left(\frac{5\sqrt{2}}{8}\right)^2 + \left(\frac{5\sqrt{2}}{4}\right)^2}\) \(|\mathbf{E}_{P}| = \sqrt{\frac{50}{64} + \frac{50}{16}}\) \(|\mathbf{E}_{P}| = 2.5\sqrt{2}\)
03

Find a unit vector in the direction of \(\mathbf{E}\) at point \(P\)

To find a unit vector in the direction of \(\mathbf{E}\) at point \(P\), divide the vector \(\mathbf{E}_{P}\) by its magnitude: \(\mathbf{\hat{E}}_P = \frac{\mathbf{E}_{P}}{|\mathbf{E}_{P}|} = \frac{1}{2.5\sqrt{2}} (\frac{5\sqrt{2}}{8}\mathbf{a}_r + \frac{5\sqrt{2}}{4}\mathbf{a}_{\phi})\) \(\mathbf{\hat{E}}_P = \frac{1}{5} (\mathbf{a}_r + 2\mathbf{a}_{\phi})\) So, the unit vector in the direction of \(\mathbf{E}\) at point \(P\) is: \(\mathbf{\hat{E}}_P = \frac{1}{5} \mathbf{a}_r + \frac{2}{5}\mathbf{a}_{\phi}\)

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Most popular questions from this chapter

A field is given as \(\mathbf{G}=\left[25 /\left(x^{2}+y^{2}\right)\right]\left(x \mathbf{a}_{x}+y \mathbf{a}_{y}\right) .\) Find \((a)\) a unit vector in the direction of \(\mathbf{G}\) at \(P(3,4,-2) ;(b)\) the angle between \(\mathbf{G}\) and \(\mathbf{a}_{x}\) at \(P\); (c) the value of the following double integral on the plane \(y=7\). $$ \int_{0}^{4} \int_{0}^{2} \mathbf{G} \cdot \mathbf{a}_{y} d z d x $$

Given the points \(M(0.1,-0.2,-0.1), N(-0.2,0.1,0.3)\), and \(P(0.4,0,0.1)\), find \((a)\) the vector \(\mathbf{R}_{M N} ;(b)\) the dot product \(\mathbf{R}_{M N} \cdot \mathbf{R}_{M P} ;(c)\) the scalar projection of \(\mathbf{R}_{M N}\) on \(\mathbf{R}_{M P} ;(d)\) the angle between \(\mathbf{R}_{M N}\) and \(\mathbf{R}_{M P}\).

Given the vectors \(\mathbf{M}=-10 \mathbf{a}_{x}+4 \mathbf{a}_{y}-8 \mathbf{a}_{z}\) and \(\mathbf{N}=8 \mathbf{a}_{x}+7 \mathbf{a}_{y}-2 \mathbf{a}_{z}\), find: ( \(a\) ) a unit vector in the direction of \(-\mathbf{M}+2 \mathbf{N} ;(b)\) the magnitude of \(5 \mathbf{a}_{x}+\) \(\mathbf{N}-3 \mathbf{M} ;(c)|\mathbf{M}||2 \mathbf{N}|(\mathbf{M}+\mathbf{N})\)

Point \(A(-4,2,5)\) and the two vectors, \(\mathbf{R}_{A M}=(20,18-10)\) and \(\mathbf{R}_{A N}=(-10,8,15)\), define a triangle. Find \((a)\) a unit vector perpendicular to the triangle; \((b)\) a unit vector in the plane of the triangle and perpendicular to \(\mathbf{R}_{A N} ;(c)\) a unit vector in the plane of the triangle that bisects the interior angle at \(A\).

Express the uniform vector field \(\mathbf{F}=5 \mathbf{a}_{x}\) in \((a)\) cylindrical components; (b) spherical components.
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