Safari Guide: Irvin |
OPENING QUESTIONS: What are some of the larger topics covered in E&M? Because a FRQ has many parts, what are some skills and topics that are linked together?
OBJECTIVE:
WORDS/FORMULAE FOR TODAY:
TERMS:
CONSTANTS:
FORMULAE:
WORK O' THE DAY: Alrighty safari tourists, here is where I put on my guide hat and lead you along the way!
Let’s start with an overview on the problem: The figure shows two point charges, one with a defined charge and another with an unknown charge, and their electric field lines. The graph gives us the electric potential along equipotential lines and it is given each box is 0.5 m wide. The word “equipotential” is specifically referenced here, so remember from the terms section earlier that equipotential means the electric potential is equal along the entirety of that line.
Immediately, we should recognize that topic area(s) that are being covered. At the minimum, we are likely to be tested on charge and volts, with the possibility of Gauss as well. I like to have a general understanding of the topics being tested before I jump into the question. It could help to identify the section in the formula sheet that covers these units.
Now that we have a solid overview of the question, let’s jump into part a. Weeeeeeee!
Part a asks us to find the charge of the unknown charge, very predictable. We have the initial conditions of one charge and electric potential. What formula relates the two? Well, we can use the formula V= (kq)/r ! But wait, we don’t have r. Remember that the distance r is 0.5 m across any grid, so we actually do have the distance r between points. Now, we should realize electric potential along one line is the result of the pull of both points, so we add the two (kq)/r of both points to get the total V of one line. What line would make relating the two points easiest? I would say the line with a electric potential of 0 V because then we can write an equation 0=V_{1}+V_{2}. This is the first point. Then, we can set -V_{1}=V_{2}. This is really important because we already have all the information for one point, and we can use basic algebra to solve for the missing charge of the other point. Plug in (kq)/r for V. We can cancel out the k constant as it appears on both sides. We’re left with -q_{1}/r_{1} = q_{2}/r_{2}. In order to find the radii, count the number of grids between the points and the line. This equals 2.5m for point 1 and 1m for point 2. Now, we can just plug in and solve. -q1/2.5 = 2.0nC/1 So, q1= -5.0nC
Part b asks us to label the direction of the electric field line at point C on the diagram. What direction does an electric field line point in? Well, electric field lines point towards regions of lower electric potential. At point C, the electric potential is -12V. Notice the equipotential line directly below has an electric potential of -16V, which is less than -12V while the other side has a line of -8V. So, the electric field line would point directly perpendicular to the -12V line and towards the -16V line because it is the closest path to the lower electric potential. Easy when you think about it!
Part c here we go! The question asks us to find the electric field strength. The only data we have is electric potential here, so let’s use a formula that relates the two. E_{x} = -dV/dx looks like a good one because it directly relates the two. However, we don’t have the equation for either so we can’t take infinitely small slices. We can, however, approximate on a “macro” scale by using E_{x}=∆V/∆x with the graph. Take the absolute value because the change in potential doesn’t regard the sign. The ∆V between point D is -20-(-24) so +4 and the distance between the two lines is 0.2m. This stems from each grid being 0.5m with 5 smaller boxes in each grid so each box is 0.1m. The distance between the lines has two boxes so 2 x 0.1m. In all, 4/0.2 gives us an answer of 20N/C.
Part d is a simple Gauss. Take Gauss’s law of Φ = qin/ε_{o} because it gives us electric flux. The equipotential of 0V encloses the charge of 2.0nC so that will be qin. Plug in the value for the vacuum permittivity and we get:
Part e part i asks for the total work done by a proton moving across the figure. Remember that we are mainly working with electric potential and the only equation relating work and voltage is: W=-q∆V Because it’s a proton, plug the charge of a proton in for “q”. The value of ∆V is more difficult, but it gives us the boundaries from A to E. The change in V from these points is from 4V to -4V, or -8V when doing final minus initial. So, the formula is: W=-1.60 x 10-^{19}x 8 which is 1.28 x 10^{-18} J
Part e part ii asks us for the speed, or kinetic energy, of said proton. The basic energy of kinetic energy relates our answer from part i to what we are looking for here. W=KE=1/2mv^{2} . So substituting in from the previous example and the mass of a proton (note: most times in a two part question, the answer from the first part will be used in the second), we get. Therefore: v= 3.92 x 10^{4} m/s
Finally, we arrive at part f. Part f asks what direction an electron would accelerate if it were placed at point B. What direction do electrons accelerate? Well, it depends on the context. Most notably, the direction of the electric field line as electrons accelerate in the opposite direction. The electric field line at point B would point to the right as the electric potential to the right is less, so the electron would accelerate to the left. Rephrasing my explanation above is a good justification.
We’re done! Weeeeeeeee!
I used the 2016 FR #1 |