The horizontal axis is the ratio of the distance to the rod divided by the length of the rod. Number of pieces (for the numerical calculation) = 100.I want to plot the magnitude of the electric field as a distance from the rod for all three methods (the two equations and the numerical method). The second formula is an approximation if the length of the rod is long compared to the distance from the rod. Skipping the derivation, I have two expressions for the magnitude of the electric field along an axis perpendicular to the center of the rod. This is a region that I can also calculate the electric field using calculus such that I can see how well the two methods agree. In order to determine the accuracy of this numerical model, I need to calculate the electric field along an axis perpendicular to the rod and in the center of the rod. That looks very pretty, but it's not that useful. However, I will show you what it looks like. There are probably many introductory physics classes that use this problem as part of a homework assignment or something. I know, that sort of stinks - but that's the way things are going to be. The total electric field is the vector sum of the two parts.Now for the program. Or an updated version of the article here: The second asymmetric line looks a lot like a similar example problem we did about half way through this article: The first symmetric line we already solved in this article. If you cut the line at 6 cm you get a symmetric problem (6cm line with q at 3cm) plus an asymmetric chunk off to one side (4cm line with q positioned 3cm off the end). If you want to solve the harder problem of an asymmetric line, (example: a 10 cm line, with q 1cm away opposite the 3cm point) you can break the line up into two sections. If you are standing at 40cm and turn your head to look at one end, you pretty much turn 90 degrees. It assumes the angle looking from q towards the end of the line is close to 90 degrees. The long line solution is an approximation. The solutions presented here assume the test charge q is straight across from (or near) the middle of the line of charge.įor a long line (your example was 1cm away from a 100cm line), the test charge q should be somewhere in the vicinity of the 50cm mark on the line, say something like +/- 10cm. It should look like a sine curve, but with a twist. Draw a third graph underneath the second one. What you should get should look like a cosine curve. Do that for a bunch of different x values. On the lower graph, plot a number you guess for that slope at coordinate (x=0, y=somenumber). ![]() Replace the q with d q, take the center of the rod as origin, and the rod as the x-axis. ![]() For each point along the sine curve, plot the slope on the graph underneath.įor example, at x = 0, sin x = 0, and the slope is steep. Use the conical flux formula where for a charge at the tip of a cone, the flux through the base of a cone with semi vertex angle due to a charge at the top of the cone is. Then set up a separate graph a little lower on the page, with the y axis lined up with the sine graph. One problem is Sal just gives the answer without showing why.Ī great exercise to understand the derivative of sine is to draw a sine wave on paper. He covers the derivative of sine and cosine here. Integration is the reverse process of differentiation, so "anti-derivative" is an appropriate term. (He refers to it as the "anti-derivative" of cosine, which means the same thing.). E y = 1 4 π ϵ 0 μ a sin θ ∣ − π / 2 + π / 2 = 1 4 π ϵ 0 μ a ( + 1 − − 1 ) = 2 4 π ϵ 0 μ a \displaystyle E_y = \dfrac E y = 4 π ϵ 0 1 a μ sin θ ∣ ∣ ∣ ∣ ∣ − π / 2 + π / 2 = 4 π ϵ 0 1 a μ ( + 1 − − 1 ) = 4 π ϵ 0 2 a μ E, start subscript, y, end subscript, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, mu, divided by, a, end fraction, sine, theta, vertical bar, start subscript, minus, pi, slash, 2, end subscript, start superscript, plus, pi, slash, 2, end superscript, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, mu, divided by, a, end fraction, left parenthesis, plus, 1, minus, minus, 1, right parenthesis, equals, start fraction, 2, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, mu, divided by, a, end fractionĬheck out Sal's video on the integral of cosine.
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