The FYS1120 Weekly Assignment Post #2

This post is from an earlier year, meaning the information here is likely outdated. You should look for a newer post from the current year to get the newest exercises and notes.

Allright! Here are my solutions to this weeks assignment. If you have any questions just post them here and someone might just help you out. The same goes if the solutions are not in agreement with your own.

Assignment #6

a) $E =\frac{E_0}{\kappa_1} \ \ 0 < y < d/2 \ \ \, E = \frac{E_0}{\kappa_2} \ \ \ d/2 < y < d$

$\Delta V = \int_0^d \vec{E} \cdot \vec{dy} = \frac{E_0 d}{2 \kappa_1} + \frac{E_0 d}{2 \kappa_2} = \frac{E_0 d}{2} \left( \frac{ \kappa_1 + \kappa_2}{\kappa_1 \kappa_2}\right) = \frac{Qd}{A\epsilon_0 2} \left( \frac{ \kappa_1 + \kappa_2}{\kappa_1 \kappa_2}\right)$

b) $C = \frac{Q}{\Delta V} =\frac{2Q}{E_0 d}\left(\frac{\kappa_1 \kappa_2}{\kappa_1 + \kappa_2}\right) = \frac{2A\epsilon_0}{d}\left(\frac{\kappa_1 \kappa_2}{\kappa_1 + \kappa_2}\right)$

c) $C = \frac{A \epsilon_0}{2d} (\kappa_1 +\kappa_2)$

a) $R = \frac{\rho}{A_0 \alpha} \ln(1+L\alpha)$

b) L’Hopital

c) Hmm.. not yet sure.

a) $E = \frac{D}{\epsilon_0 \kappa} = \frac{Q}{4\pi \epsilon_0 \kappa} \frac{1}{r^2}$

b) $C = \frac{Q}{4 \pi \epsilon_0 \kappa} \left(\frac{b-a}{ab}\right)$

c) $Q_a = Q_b \Rightarrow \sigma_b = (\frac{a}{b})^2 \sigma_a$

d) $\sigma_a = \hat{n}\cdot \vec{P} = \sigma \frac{\kappa -1}{\kappa} \ \ \ r=a$

$\sigma_b = \sigma \frac{a^2}{b^2} \frac{\kappa - 1}{\kappa} \ \ \ r=b$

a) $I_1 = \frac{12}{11}A$ , $I_2 = \frac{8}{11} A$ , $I_3 = \frac{4}{11} A$

b) $P = I\epsilon = 8.72 W$

c) $P = P_1 + P_2 + P_3$

4 thoughts on “The FYS1120 Weekly Assignment Post #2”

Another suggestion for 2c:
J = I/A. So if you’re going to make J r dependent, you either have to do something about A or something about I. Changing I wont affect R, so you’re left with doing something about A. But A varies only with x, so there would be no change.

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A suggestion for 2c:
If the current density is a function of r, $J = J(r)$ for x=const, the resistivity is not constant in the radial direction, or $\rho = \rho(r)$ . I think this would give a different result since R is dependent on the resistivity. If the resistivity somehow remains constant, the result should be as in 2a (or if the total integrated resistivity equals $\rho \ \forall \ x$ ).
This may or may not have anything to do with the correct answer :p