<p>E&M 1:
a) Suppose that the shell of charge has radius R. Construct a Gaussian sphere of radius r such that r<R. Because this sphere encloses no charge, the net flux through it is zero. Because of radial symmetry, the field is the same everywhere and is therefore exactly equal to the total flux divided by the area of the sphere. Since the total flux is zero, the field everywhere within the shell must be zero.</p>
<p>b) No. Though Gauss' Law can be used to prove that the net flux through any Gaussian sphere of radius r<R is zero,but it cannot be used to compute the field, because the field does not follow radial symmetry. The E term in the closed integral E dot dA cannot be taken out of the integral for this reason and therefore the field may have non-zero values at points on the surface, provided the total flux is zero. There is, therefore, no reason for the field to be zero everywhere, since the charge distribution lacks symmetry.</p>
<p>c) ABCD ABGH ADEH
Basically, all the square surfaces that included A as a surface have zero flux, since the field on those points is parallel to the plane itself. Flux counts only the component of the field that is normal to the area. In these cases, the component of the field normal to the surface is zero, so the charge contribute no flux through these surfaces.</p>
<p>d) F
Basically, for the electric field due to a point charge to be minimized, you need to be as far away from the charge as possible. Choice F is the farthest and is therefore the correct answer.</p>
<p>e) The distance from the charge is the square root of L^2+L^2+L^2 = 3L^2
The field then = kQ/(3L^2). You could use k or you could use epsilon naught.</p>
<p>f) Gauss' Law is the fastest way to calculate flux. The enclosed charge is Q/8. The flux coming out of each square is the total flux divided by 6, since the cube has 6 faces. The total flux is Q/(8<em>epsilon). So the flux through that one square is Q/(48</em>epsilon). </p>
<ol>
<li>a)
i) Use the equation q=CV. Adjust for units if necessary. The voltage is the voltage of the battery and the capacitance is that of the capacitor provided.</li>
</ol>
<p>ii) At t1 the charge is zero. As time approaches infinity, the charge builds to CV. It never reaches CV though. It just approaches it as a horizontal asymptote.
The shape of the function is similar to that of 1-e^(-t) which happens to asymptotically approach 1. </p>
<p>iii) Same as the previous part except instead of CV the ideal current is V/R. </p>
<p>b) i) You merely had to use the equation E= (q^2)/2C. q is given in the equation. C is given in the circuit diagram. Adjust for units as necessary.</p>
<p>ii) Use the equation (q^2)/2C = (1/2)L(I^2). You know q, you know C, you know L. You need to solve ofr I.</p>
<p>iii) I got this part wrong, but now I believe the answer is:
V - LI’ - q/c = 0
You know V, you know L, you know q, you know C. Solve for I’</p>
<p>3)
a)
i) B = 0 because the inside section encloses no current.</p>
<p>ii) Use the proportion of area to find the value of the actual current enclosed by the ring. Then use ampere’s law applying symmetry to a circular Amperian loop to solve for B.</p>
<p>iii) Same as above but you don’t need to do the proportion for area, since you already know the effective current enclosed by the Amperian loop.</p>
<p>b) It is pointing down-right relative to you looking straight down at the page.</p>
<p>c) No magnetic force because the charge is not moving (F=qvb=0 since v=0).</p>
<p>But if there is no flux through those three surfaces, doesn’t all of the flux go through the other three surfaces? The total flux has to be equal to Q/(8ε) - and we determined in part (c) that there is no flux through surfaces ABCD ABGH ADEH because the E-fields are parallel to their areas - so all of the flux must go through the remaining three surfaces, making the flux through CDEF (Q/(8ε))(1/3) = Q/(24ε)</p>
<p>My Mechanics Solutions, copied from another thread:</p>
<p>My procedure/thought process for #1 Mech:</p>
<p>For (a) and (b), v is v_x</p>
<p>(a) J=Ft, so t=J/F
(b) J=mv, so m=J/v
(c) Work done in stopping the projectile is the change in kinetic energy. KE=.5mv^2, so it’s .5(J/v)v^2, which is .5Jv
(d) Work is equal to average force times distance. So .5Jv=Fd. So F_b=.5Jv/d
(e)/(f) I know that the impulse imparted by the projectile on the block is equal to something, which I wrote on the test for both parts. Hopefully I get points.</p>
<p>Same for #2 Mech:</p>
<p>(a) 2 forces, normal and gravity. Gravity straight down, Normal perpendicular to surface
(b) Screwed this one up, said Mgcos(theta). Maybe I’ll get something for writing normal=centripetal?
(c) Conservation of energy from point A (all gravity potential) to point D (all kinetic). h for mgh is 7R/4.
(d) I only remember getting 7/12 for my coefficient of friction. Which might not even be what I put.
(e) i. -kv=m(dv/dt)
ii. Integrate
iii. Initial acceleration (i.e. at point D) is kv/m (even though I only put v). Concave up, decreasing, horz. asy. of acceleration equals 0.</p>
<p>For Mech. #3 (a=alpha, B=beta, O=theta):</p>
<p>(a) Ia=-BO. Replace a with double derivative of theta with respect to time and divide I over.
(b) Using small angle approximation, alpha=-(angular velocity)^2 times theta. Angular velocity is sqrt(B/I). Period=2pi/angular velocity. Could also use mass on spring analogy, with m=I and k=B.
(c) Graph.
(d) My slope was like 160, y-int .20. y=160x+.2 or T^2=160I+.2
(e) Checked my calculator, got .247. Units are same as that of spring constant.
(f) No clue.</p>
<p>My E&M solutions, again, copied from another thread:</p>
<p>Anyway, my answers. Some are definitely wrong:</p>
<p>E&M. 1.</p>
<p>(a) Surface is sphere with r<R and goes inside shell. Basically, the surface encloses no charge whatsoever. By Gauss’ Law, if no charge is enclosed, electric field must be 0.
(b) Yes. While the charge density on the outside of the sphere is not uniform, the Gaussian surface still resides within the shell, in which no charge is enclosed, so E is still 0.
(c) I said none, since I reasoned that no matter what, it would be impossible for all field lines to be parallel to the face. But I agree that it’s all faces with A
(d) Corner F
(e) E=(kQ)/(3L^2)
(f) I did some really random stuff, integrating E dA. I said A=L^2, so dA=2LdL. I subbed that in, said that E=(kQ/8L^2) and I integrated from L to L*sqrt(3). Definitely got something weird. Hopefully I get points.</p>
<p>E&M. 2.</p>
<p>(a) i. Q=CV. Make sure units of capacitance are farads and not mF.
ii. Q=0 from t=0 to t=t<em>1. Increases exponentially (concave down) with a horizontal asymptote of CV (or equivalent number)
iii. I=0 from t=0 to t=t</em>1. I jumps up to V/R (or equivalent number) at t=t_1 and decreases exponentially (concave up) with horz. asy. of I=0.
(b) i. U=.5QV. Modify so that you have Q and C and solve (I may have plugged in 9 for V without thinking…ugh)
ii. Conservation of energy. The energy from part (b)i equals .5LI^2, with L being the inductance. Solve for I (I got .297 A).
iii. Use V=Q/C to solve for voltage. V=-L(dI/dt). Solve for dI/dt.</p>
<p>E&M. 3.</p>
<p>(permittivity constant/2pi)=(2E-7)</p>
<p>(a) i. B=0, since dl is zero since no current flows in that section.
ii. B=(2E-7)(I)(r^2/b^2)
iii. B=(2E-7)(I/4b)
(b) I put that it went horizontally to the right, though it should be pointing right and down.
(c) No force. No electric field is present, so no electric force. For magnetic force to be felt, electron must be moving (F=qvB). However, v=0, so magnetic force=0.
(d) i. Graph
ii. Plug in pi, b, and I values from directions. Get B/r from the slope of line. Solve for permittivity constant (I got something like 1.46E-7; it was very close to the real value).</p>
<p>Ok so I still think that 1d. is A, because there’s no charge (and hence, no field) at the center of a conducting sphere (so A has E = 0, and is the least).</p>
<p>For 1f. I think the answer is Q/24epsilon. Imagine that you have 8 identical cubes with one vertex at A (the one shown, and 7 other cubes). This way, we can treat the charge as a point charge within the giant cube created by the 8 cubes. EdA=Q/epsilon, so we then just divide (Q/epsilon) by the number of identical squares there are, which is 24 (6 big square sides of the big cube, 4 smaller sized sides per big side, so 24 small sides). The flux through each of these 24 faces will be identical, so one side (the one listed) will have flux=Q/24epsilon.</p>
