A group of students trace out two equipotential surfaces and an electric field l

Question

A group of students trace out two equipotential surfaces and an electric field line as shown in Figure 1.5-1. A) Identify the polarity of each terminal. B) Calculate the electric field for each segment of the electric field line given that the distance between the two digital multi-meter probe tips is 2 times 10^-2 m. C) Determine the potential difference between the two equipotential lines by summing the voltages along the electric field line. Compare this value to the potential difference found by subtracting the voltages of the two equipotential lines.

Explanation / Answer

The terminal with higher electric potential is positive and the one with lower electric potential is nehative, since current flows from higher electric potential to lower electric potential or from positive terminal to negative terminal.2 volt is the negative terminal and 8 volt is the positive terminal.

Electric field E = potential/distance

1) 1.9/2X10-2=95V/m

2)1.2/2X10-2=60V/m

3)0.6/2X10-2=30V/m

4)1.1/2X10-2=55V/m

5)1.5/2X10-2=75V/m

Potential difference from subtracting the two equipotential lines= 8-2= 6V

By adding the potentials between the two equipotential lines we get =1.9+1.2+0.6+1.1+1.5 = 6.3 V

There is a diffrence of 0.3 V between the two values.

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