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Tutor profile: Daynah R.

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Daynah R.
Current graduate student at the University of Central Florida pursuing an Applied Photonics Grad Certificate
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Questions

Subject: Physics (Waves and Optics)

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Question:

The complex time harmonic electric field of a plane wave $$(\lambda_0$$ = $$0.2 \pi$$ microns) propagates in a $$\textbf{lossless}$$ dielectric medium. Determine the refractive index of the medium and the E-field component in the y-direction. $$ \textbf{E} = \pi(96 \hat{x} + E_y \hat{y})$$ $$e^{-jk_0(1.5 \textbf{x}+2\textbf{y})}$$ V/m

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Daynah R.
Answer:

To find the refractive index of the medium, we recall that $$ \textbf{k} \cdot \textbf{k} = k_0^2 n^2$$ k being the magnitude of the propagation vector and $$k_o$$ the plane wave propagation $$k_o = \frac{2 \pi}{\lambda_0}$$ =$$ \frac{2 \pi}{0.2 \pi \mu m}$$ = 10 $$\mu m^{-1}$$ We are given the propagation vector, which is: $$ \textbf{k} = k_0(1.5\textbf{x}+2\textbf{y})$$. With this, we take the dot product of the propagation vector $$ \textbf{k} \cdot \textbf{k} = k_0^2 2.25+k_0^2 4 $$ = $$k_0 6.25$$ Now we can find the index of refraction $$ \textbf{k} \cdot \textbf{k} = k_0^2 n^2$$ $$ ({10 \mu m^{-1}})^2$$ $$6.25 =$$ (10 $$\mu m^{-1})^2$$ $$n^2$$ Solving for n, we get n = 2.5 Next, we need to find the E-field component in the y-direction. $$ \textbf{k} \cdot \textbf{E} $$ = 0 $$ [k_0(1.5\textbf{x}+2\textbf{y})] \cdot [\pi(96 \hat{x} + E_y \hat{y})$$ $$e^{-jk_0(1.5 \textbf{x}+2\textbf{y})}]$$ = 0 Solving for $$E_y$$, $$ [(1.5\textbf{x}+2\textbf{y})] \cdot [(96 \hat{x} + E_y \hat{y})]$$ = 0 $$144 + 2 E_y$$ = 0 $$E_y$$ = -72

Subject: Physics (Electricity and Magnetism)

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Question:

Find the electric field outside of a uniformly charged solid sphere of radius R and total charge q.

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Daynah R.
Answer:

First, imagine a solid sphere of radius R. Then a Gaussian sphere of radius r > R. Using Gauss's Law: $$\oint_S \textbf{E} \centerdot \textit{d}\textbf{a} = \frac{1}{\epsilon_0} Q_{Enc}$$ Since we are looking for the electric field outside of the surface, $$Q_{Enc} = q$$. In this case, $$\textbf{E}$$ and $$\textit{d}\textbf{a}$$ point radially outward, and so the dot product can be dropped. The magnitude of $$\textbf{E}$$ is constant over the Gaussian surface. $$\int_S \textbf{|E|}$$ $$\textit{da}$$ =$$ \frac{1}{\epsilon_0} q$$ Recall the area of a sphere. $$\textbf{E}$$ $$4\pi r^2$$ =$$ \frac{1}{\epsilon_0} q$$ $$ \textbf{E} = $$ $$ \frac{q}{{4\pi r^2}{\epsilon_0}} {\hat{r}}$$ Here, we have found the electric field outside of the sphere. Note that the electric field here is a vector and so we must show the vector in spherical coordinate.

Subject: Pre-Calculus

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Question:

Find the inverse of $f(x) = sqrt{x+8}$. Verify your answer.

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Daynah R.
Answer:

For $$f(x) = \sqrt{x+8}$$, Replace f(x) with y: $$y = \sqrt{x+8}$$ Then, switch the position of x and y: $$x = \sqrt{y+8}$$ Now, solve for y to find the inverse of the function: $$x = \sqrt{y+8}$$ $$x^2 = y + 8$$ $$y = x^2 - 8$$ Which is now $$f(x)^{-1 } = x^2 - 8$$ We have found the inverse of the function, but we need to verify it to see if it's correct. We can verify it by using $$(f^{-1} \circ f) (x) = x$$ $$(f^{-1} \circ f) (x) = f^{-1} [f(x)]$$ $$= (f(x))^2 - 8$$ Remember that $$f(x) = \sqrt{x+8}$$ $$= (\sqrt{x+8})^2 - 8$$ $$=x + 8 -8$$ $$ = x$$, and so ($$f^{-1} \circ f) (x) = x$$ is true

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