BW#8 Ch 21: P. 20 – 24
P21.20. Prepare: Please refer to Figure P21.20. In a region that has a uniform electric field, Equation 21.17 gives the
magnitude of the potential difference between two points.
Solve: (a) The electric field points “downhill.” So, point A is at a higher potential than point B.
(b) The magnitude of the potential difference between points A and B is
V  Ex d  (1000 V/m)(0.07 m)  70 V
That is, the potential at point A is 70 V higher than the potential at point B.
Assess: Electric field points from higher potential to lower potential.
P21.21. Prepare: In order to use Equation 21.17, E  V / d , we need to know the shortest distance between two
equipotential surfaces. We are given the distance between point A which is on one surface and point B which is on another
surface. However this is not the shortest distance, as you can see from the figure. The shortest distance, d , is given by
d  (5.0 cm)cos30  2.5 3 cm.
Solve: Since the electric field vectors point from the surface containing A to the surface containing B, A is at a higher
potential and VA  VB will be a positive potential difference. From Equation 21.17, the potential difference is given by:
VA  VB  Ed  (1200 V/m)(2.5 3 102 m)  52 V
Given that VA  300 V, we conclude that VB  352 V.
Assess: Even though the electric field vectors do not point directly from point A to point B, the potential is less at B than
at A because the displacement vector from A to B forms an acute angle, 30 , with the electric field. That means that to
get from A to B, one component of your motion must be in the direction of the field.
P21.22. Prepare: The magnitude of the electric field in terms of the potential difference V between two points a
distance d apart is given by Equation 21.17.
Solve: When the electric field is uniform,
V  Ex d  Ex x  (1000 V/m)(0.30 m  0.10 m)  200 V
P21.23. Prepare: Please refer to Figure P21.23. The three equipotential surfaces correspond to potentials of 0 V, 100
V, and 200 V. The electric field points in the direction of decreasing potential and is perpendicular to the equipotential
lines.
Solve: The magnitude of the electric field between equipotential surfaces a distance d apart is
E
V 100 V

 10,000 V/m
d
0.01 m
The direction of the electric field vector is “downhill,” perpendicular to the equipotential surfaces, and directed to the left.
That is, the electric field is 10,000 V/m to the left.
P21.24. Prepare: Please refer to Figure P21.24. Three equipotential surfaces at potentials of 200 V, 0 V, and 200 V
are shown. The electric field is perpendicular to the equipotential lines and points “downhill.”
Solve: The electric field perpendicular to the equipotential surfaces is
E
V 200 V

 20,000 V/m
d
0.01 m
The electric field vector is in the third quadrant, 45 below the negative x-axis. That is,
Assess:
E  (20,000 V/m, 45 below  x-axis)
As obtained previously, electric field points from higher potential to lower potential.