00:01
In this problem on the topic of electric potential, we have an electric field with a constant value of 4 times 10 to the 3 volts per meter directed downward.
00:08
The field is the same everywhere, and the potential at p within this region is 155 volts.
00:15
We want to find the potential at 6 .6 times 10 to the minus 3 meters directly above p, 3 times 10 to the minus 3 meters directly below p, and then 8 times 10 to the minus 3 meters directly to the right of p.
00:28
Now the drawing here shows the electric field e and the three points a, b, and c in the vicinity of point p, which would take us the origin.
00:36
We choose the upper direction as being positive, and so e is minus 4 times 10 to the power 3 volts per meter.
00:52
Now, the electric potential at points a and b can be determined from equation 7 as delta v is equal to, minus e delta s, since e and delta s are known.
01:08
Since the path from p to c is perpendicular to the electric field, no work is done in moving a charge along such a path.
01:14
Thus, the potential difference between these points is zero.
01:18
And so for part a, the potential difference between p and a is va minus vp, which is minus e delta s...