Download Example 17-4 Electric Potential Difference in a Uniform Field I

Example 17-4 Electric Potential Difference in a Uniform Field I
A uniform electric field points in the positive x direction and has magnitude 2.00 * 102 V>m. Points a and b are both
in this field: Point b is a distance 0.300 m from a in the negative x direction. Determine the electric potential difference
Vb 2 Va between points a and b.
Set Up
We apply Equation 17-7 to the path shown
in the figure. Note that the magnitude of the
displacement is d = 0.300 m, and the angle
between the electric field and the displacement
is u = 180°. Note also that the potential difference V equals the potential at the end of the
displacement ds (that is, at point b) minus the
potential at the beginning of the displacement
(that is, at point a).
Solve
Use Equation 17-7 to solve for the potential
difference.
Potential difference between
two points in a uniform
electric field:
V = Vb 2 Va = 2Ed cos u
E = 2.00 × 102 V/m
d = 0.300 m
(17-7)
b
a
Calculate the potential difference from the electric field magnitude E,
the displacement d, and the angle u:
V = Vb 2 Va = 2Ed cos u
= - 12.00 * 102 V>m2 10.300 m2 cos 180
= 2(60.0 V)(21)
= +60.0 V
Reflect
We can check our result by comparing with
­Example 17-1 in Section 17-2, where we
­considered the change in electric potential
energy Uelectric for an electron that undergoes
the same displacement in this same electric
field. Using Equation 17-6, we find the same
value of Uelectric as in Example 17-1.
The positive value of V = Vb 2 Va
means that point b is at a higher potential
than point a. This agrees with our observation
above that if you travel opposite to the direction of the electric field, the electric potential
increases. The value E = 2.00 * 102 V>m
means that the electric potential increases by
2.00 * 102 V for every meter that you travel
s.
opposite to the direction of E
+x
If a charge q0 = 21.60 * 10219 C travels from a to b, the charge in
electric potential energy is given by Equation 17-6:
V =
Uelectric
q0
so
Uelectric = q0 V = (21.60 * 10219 C)(+ 60.0 V)
= 29.60 * 10218 V # C = 29.60 * 10218 J
(Recall from above that 1 V = 1 J>C.)