Table 2 Two Parallel Line Charge Configuration X (m) 0.020 0.040 0.060 0.080 0.1
ID: 1577183 • Letter: T
Question
Table 2 Two Parallel Line Charge Configuration X (m) 0.020 0.040 0.060 0.080 0.100 0.120 0.140 0.160 0.180 E (V/m) 4.99 4.20 2.44 20 0.B7S 3. Complete the chart by calculating the average magnitude of the electric field between each two locations by dividing the difference in the magnitude of the voltages at each of the consecutive two locations by the displacement between the two locations. Then find their average value and record it in the last row. (10 points) a ae-430 3+v/m = Calculate the average value of the electric field along the straight line between the two parallel line charges by dividing the difference in the magnitude of the voltages at the two parallel lines by the displacement between them. Then take the % difference between your two values for the average electric field between the parallel line charges. (5 points) 4. A straight electric field line is supposed to represent a constant electric field. Does your data for both configurations, more or less, agree with this? If not what are some plausible explanations? (5 points) 5.Explanation / Answer
3. from the chart we can see
E avg = 30.875 V/m
4. now, from the chart consider the first two values of the electric field
E1 = 34 V/m
% difference= |Eavg - E1|*100/Eavg = 10.121 %
E2 = 33.5 V/m
% difference = 8.5 %
5. the data in the table more or less agrees with the fact that the electric field magnitude is fairly constant throughout the length of the line. there are a few places where there are outlier reaodngs , ,but these can be due to manual error and random errors as well
6. the data is consistent for line field configuration than for two piint charge configuration because two point charge configuration follows inverse square law for variation of electric field, which is not constnt with distance and hence it is an entirely different vcase and hence the data is not consistent with that
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