Lab 11: It\'s Electric! During class, your illustrious professor collected data

Question

Lab 11: It's Electric! During class, your illustrious professor collected data and linearized a graph for a second problem statement, "How does electric force (FE) depend on charge (Q or q)?" In case you need it, the simulation is provided here: http://tube geogebra.org/student/m499903 1. (0.5 pts) Write the specific math model that was recorded for this problem statement. Also, write the value of each variable held constant during the experiment. 2. (1.5 pts) By combining all three problem statements, we determined the general math model: kQq Using this model and the slope from the problem above, determine the valtue of constant k. Note: did this in class for your Problem Statement #1. uidntit ih maed rnd u oe rtoat hre prehlbweermine thease of constant k Note you At right is a set of data collected for two electrically charged conducting spheres. The r (m) F CN) heres are brought closer and closer together, reducing the separation (r) between them. As with Lab 11, this data will allow us to answer the problem statement: How does the electric force depend on the separation between charges? 0.675 0.5 2.7 10.8 30 3. (1 pts) Using Logger Pro, plot these variables and linearize the graph. Write the te the 0.25 specific math model here: 0315 0.15 0.1 0.05 67.5 270 0.025 1080 Determine the charge on each sphere, utilizing the slope from above and the 0015 3000 4. (1 pt) One of the spheres is carrying three times as much charge as the other known general math model (Coulomb's Law). 0.01 6750

Explanation / Answer

For Problem 1 , The math model is FE r2 = 0.675 ( I gues) You need to provide data like in problem 3

This holds for every value. The variables held constant are both charges q and Q

Problem 3 is correct

Problem 2 we need the values of q and Qfor which the experiment was done.

So Slope = F * r^2   = 0.675 in problem 3

This will be equal to k *q * Q

So k = 0.675 / (q * Q)

In problem 4

We know from coulumbs law k = 9* 10^9 and Q = 3q

So q = 5 * 10^-6 = 5 micro coulumb

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