Analysis Table 1 Two Point Charge Configuration E (V/m) 0.31p 0.020 0.040 0.060

Question

Analysis Table 1 Two Point Charge Configuration E (V/m) 0.31p 0.020 0.040 0.060 0.080 0.100 0.120 0.140 22.5 219 -0.160 0.180 1. Complete the chart by calculating the average magnitude of the electric field between each two consecutive locations by dividing the difference in the magnitude of the voltages at each of the two locations record it in the last row. (10 points) by the displacement between the two locations. Then find their average value and v28.5 Vs = 2.19-2.42=0.32-0.021U SVs : 3.2|-2.7 q-0.42+0.02= 21 4= 3.78-3,215 0.57. 0.02= 28.5 Calculate the average value of the electric field along the straight line between the two point charges by dividing the difference in the magnitude of the voltages at the two point charges by 2.

Explanation / Answer

2. in the given data, we take the consecutive potentials , find their difference and divide it by the step size of the distance and get the electric field at that point

hence

V x E(at that location)

0.36 0.02 31.5

0.99 0.04 28.5

1.56 0.06 23

2.02 0.08 20

2.42 0.1 18.5

2.79 0.12 46

3.71 0.14 3.5

3.78 0.16 60.5

4.99 0.18

hence

Eav = 28.9375 V/m

Eg = 28.5 V/m

taking % diff

dE = (Eav - Eg)*100/Eg = 1.535 %

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