Steven U.

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Physics (Waves and Optics)

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Question:

How can many bright peaks be seen if 600nm laser light is incident normal to 330 lines per millimeter diffraction grating?

Steven U.

Answer:

For this problem, we use the condition for bright fringes produced by a diffraction grating: $d\sin(\theta)=m\lambda$. d is the slit separation in the grating, which can be found by taking the inverse of the number of lines per millimeter ($d=\frac{1}{330}mm \sim 0.003mm$). From here, we can rearrange the diffraction grating equation for $\theta$ and begin "teasing" out the answer. $\theta$ is given by: $\theta = \arcsin\frac{m\lambda}{d}$. Plugging in values of lambda and d, while varying the maxima index (m), we find: m=0, $\theta =0$ m=1, $\theta \sim 12.2^{\circ}$ m=2, $\theta \sim 25^{\circ}$ m=3, $\theta \sim 39.3^{\circ}$ m=4, $\theta \sim 57.6^{\circ}$ m=5, $\theta \sim N.A.$ (more specifically an imaginary number in the form a+bi) This tells us, that in one direction, there can only be, up to 4 bright spots. This is mirrored on both sides of the central maximum. Meaning, that 633nm laser light through this diffraction grating will produce, at most, 9 bright fringes.

Physics (Newtonian Mechanics)

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Question:

You're moving on your next big adventure! You load your one box (you're a minimalist) into the bed of a truck. The coefficient of static friction $\mu_s$ between the truck bed and the box is 0.4. If the box has a mass of 50kg, what is the largest acceleration you can reach without causing the box to slide off the back of the truck, breaking all your things?

Steven U.

Answer:

The tricky part here is creating a free body diagram for the situation. Because I can't attach images, I'll describe one such diagram as best as possible: There is a box on the truck bed, label the interface with $\mu_s=0.4$. The force of gravity on the box of course "pulls" down, with a magnitude of $F_g=mg$, with m being the mass of the box (50kg) and g being the acceleration due to gravity ($\sim 9.8\frac{m}{s^2}$). There is also the normal force from the truck bed - box interface, pushing up on the box, a label that N. Now, for the box to not slip backward, it needs to accelerate at the same rate as the truck. On my free body diagrams, I draw accelerations as a vector with a thicker line (or something similar) than force vectors. This will be represented as a thick line to the right labeled a (say the truck is accelerating the right). So, we have a complete free body diagram, kind of. We still don't have a force associated with that horizontal acceleration. The only possible way to create that acceleration would be from the friction between the box and the bed of the truck, we, therefore, add to our diagram a force arrow in the same direction as the acceleration, label it $F_s$ for the force from static friction (recall $F_s\leq \mu_sN$). We now have a free body diagram and everything we need to solve this problem (provided we recall Newton's second law: $F_{net}=ma$). Applying Newton's second law to the horizontal direction, we get: $F_s=ma$ where a is the acceleration of the truck we're looking for, m is the mass of the box, and the static force of friction is as described earlier. Rearranging for a (and substituting the rule for static friction) we find that $a\leq \frac{\mu_sN}{m}$, meaning the maximum possible acceleration is given when the inequality is equal, or $a=\frac{\mu_sN}{m}$. We would be able to solve right out, provided we knew the normal force because we don't we go in the vertical direction and apply Newton's second law. In the vertical direction, applying Newton's second law yields $N-mg=0$ (because there is no net force in this direction). This allows us to acquire a value for the normal force, specifically $N=mg \sim 500N$ (if we use $g=10\frac{m}{s^2}$). Taking this normal force to the horizontal direction, we can now find a maximum acceleration of the truck using $a=\frac{\mu_sN}{m}$. Plugging in, we find that the maximum acceleration of the truck is $a \sim 4\frac{m}{s^2}$

Physics (Electricity and Magnetism)

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Question:

A solid, conducting sphere of the radius is concentric with a non-conducting shell with inner radius b and outer radius c (with a<b<c). If the total charge ($Q_o$) on the inner sphere is uniformly distributed throughout and the charge on the shell is $-2Q_o$, determine the magnitude and direction of the electric field as a function of radial distance (r) from the center of the configuration everywhere (that is when r<a, an <r<b, b<r<c, and c<r).

Steven U.

Answer:

For this problem, we will make use of Gauss' law ($\oint \vec{E} \cdot \vec{dA} =\frac{Q_{enc}}{\epsilon_o}$). The major challenge is establishing the right side of the equation in each region, or more specifically $Q_{enc}$. Lets first analyze all areas to find $Q_{enc}$, we'll deal with Gauss' law as a whole later: (r<a): When inside the sphere we recognize that, because the charge is uniformly distributed throughout the volume, meaning we must add more charge as r approaches the radius of the sphere a. To accomplish this, we create a volume charge density $\rho$. Like any density, we can determine it by dividing the total charge on the sphere by the total volume (specifically $\rho=\frac{Q_o}{\frac{4}{3}\pi a^3}$). To then determine the volume at a specific radius from the center r (with r<a) we multiply this charge density by the volume of the sphere at r, this gives us $Q_{enc}=\rho V=\frac{Q_o}{a^3}r^3$ (again, recall r<a). (a<r<b): When between the sphere and the concentric shell (an <r<b) we enclose the full sphere, meaning that $Q_{enc}=Q_o$. (b<r<c): When inside the shell, we recall that there must be no electric field inside a conductor. Therefore, the conductor will move charges around such that the inner surface of the conductor (the surface at radius b) will contain yield a charge such that any enclosed charge is zero when conducting Gauss' law. This means, in this specific situation, the charge on the inner surface of the conducting shell will be $-Q_o$. This leaves us with an enclosed charge of $Q_{enc}=0$ when we consider radius r inside the conducting shell. (c<r): Outside the entire setup, we can simply add the entire enclosed charge, that is $+Q_o$ from the sphere, and $-2Q_o$ from the shell. Meaning the charged enclosed by a Gaussian surface outside of the entire apparatus will be $Q_{enc}=-Q_o$ (again for c<r). It is worth noting that the charge on the outer surface of the conducting sphere will be $-Q_o$, which can be understood when considering how much of the shells charge was "pulled" to the inner surface. Lets now talk about the left side of Gauss' law $\oint \vec{E} \cdot \vec{dA}$: For this portion, one needs only to consider the high symmetry introduced by the problem, specifically radial symmetry. Creating a Gaussian surface in the form of a sphere, concentric with the center of the sphere and shell exploits this symmetry. This is because, due to the radial nature of this setup, all produced electric fields will point radially inward (or outward depending on the charge enclosed). Additionally, it can be understood that the vector normal to the Gaussian surface will also be radially outward for all cases. This means that $\vec{E} \cdot \vec{dA}$ for all cases can be reduced to $\pm |\vec{E}||\vec{dA|}$ (plus when the E-field is radially outward, minus when E is radially inward). This allows us to take $|\vec{E}|$ out of the closed integral, and compute the surface area, which for all cases will be: $\oint|\vec{dA|}=4\pi r^2$, where r will be restricted depending on the region in which we're interested. Putting it all together: Recall, for all cases, we have $\pm |\vec{E}|4\pi r^2$ for the left side of Gauss' law. For now, we will drop the plus or minus, and only deal with positive values, we will determine the sign for each case individually. (r<a): Gauss' Law becomes $|\vec{E}|4\pi r^2=\frac{Q_o}{\epsilon_o a^3}r^3$. Solving for $|\vec{E}|$, we find $|\vec{E}|=\frac{Q_o}{\epsilon_o a^3}r$. This means, as we continue further from the center, while still inside the sphere, the magnitude of the electric field grows linearly with r. Because the charge enclosed $Q_{enc}$ is positive, we can say that this electric field points radially outward. (a<r<b): Within the void between the sphere and the shell, Gauss' law becomes $|\vec{E}|4\pi r^2=\frac{Q_o}{\epsilon_o}$. Rearranging for $|\vec{E}|$, we find that $|\vec{E}|=\frac{Q_o}{4\pi r^2\epsilon_o}$, which should be easily recognized as the electric field due to a point charge. Because the enclosed charge is positive, we can say that the electric field points radially outward. (b<r<c): Inside the concentric shell, the charge enclosed was $Q_{enc}=0$, this means that the entire right-hand side of Gauss' law is zero. The only way to satisfy the equation is for the electric field to be: $\vec{E}=0$ inside the conducting shell. (c<r): Outside of the entire setup, Gauss' law becomes $|\vec{E}|4\pi r^2=\frac{-Q_o}{\epsilon_o}$. Rearranging for $|\vec{E}|$, we find $|\vec{E}|=\frac{Q_o}{4\pi r^2\epsilon_o}$. NOTICE: I removed the negative sign from the charge, that is because this is a statement for the magnitude of the electric field, the negative helps us understand that the electric field points radially inward. This equation again should be recognizable as the electric field of a negative point charge.

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