# Example 17-5 Electric Potential Difference in a

```Example 17-5 Electric Potential Difference in a Uniform Field II
Determine the electric potential difference V = Vc 2 Va between points a and c in the uniform electric field of Example 17-4. Point c is a distance 0.500 m from point a, and the straight-line path from a to c makes an angle of 126.9&deg; with
respect to the electric field.
Set Up
Again we’ll use Equation 17-7 to calculate the
potential difference between the two points.
Since the displacement from point a to point
c points generally opposite to the direction of
the electric field, we expect that V = Vc 2 Va
will be positive, just like V = Vb 2 Va in
Example 17-4.
Solve
Use Equation 17-7 to solve for the potential
difference.
E = 2.00 &times; 102 V/m
Potential difference between two
points in a uniform electric field:
V = Vc 2 Va = 2Ed cos u
a
(17-7)
d = 0.500 m
O = 126.9&deg;
c
Calculate the potential difference from the electric field magnitude E,
the displacement d, and the angle u:
V = Vc 2 Va = 2Ed cos u
= - 12.00 * 102 V&gt;m2 10.500 m2 cos 126.9
= - 12.00 * 102 V&gt;m2 10.500 m2 1 -0.6002
= +60.0 V
Reflect
The potential difference between points a and c is the same as the potential difference between points a and b in Example 17-4. Equation 17-7 tells us why this should be: The potential difference V = 2Ed cos u involves the magnitude
s. In both
E of the electric field multiplied by d cos u, which is the component of the displacement ds in the direction of E
2
examples the electric field magnitude has the same value (2.00 * 10 V&gt;m), as does the component of displacement in
the direction of the electric field (20.300 m).
Another way to come to this same conclusion is to recognize that the displacement from a to c can be broken down
into two parts: a displacement in the negative x direction from a to b, followed by a displacement from b to c in the y
direction. The potential difference for the displacement from a to b is Vb 2 Va = +60.0 V as we calculated in Example
17-4; the potential difference for the displacement from b to c is Vc 2 Vb = 2Ed cos 90&deg; = 0 since that displacement is
perpendicular to the electric field. So the net potential difference between a and c is Vc 2 Va = (Vc 2 Vb) + (Vb 2 Va) =
+60.0 V + 0 V = +60.0 V.
```