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College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Im having trouble inegrating dr

Transcribed Image Text:EXAMPLE 23.6 Finding potential by integration
By integrating the electric field as in Eq. (23.17), find the potential at a
distance r from a point charge q.
Figure 23.14 Calculating the potential by integrating É for a single
point charge.
'To point b at infinity
IDENTIFY and SET UP We let point a in Eq. (23.17) be at distance r and
let point b be at infinity (Fig. 23.14). As usual, we choose the potential
to be zero at an infinite distance from the charge q.
drae
EXECUTE To carry out the integral, we can choose any path we like be-
tween points a and b. The most convenient path is a radial line as shown
in Fig. 23.14, so that di is in the radial direction and has magnitude dr.
di = îdr, we have from Eq. (23.17) de= direction
Writing . magnitose
V – 0 = V =
E; di
r? r
f îdr =
Eor
-dr
4T€0 r²
4TEor2
= 0
KEYCONCEPT The potential difference Vab = Va – V½ between
point a and point b equals the integral of the electric field along a path
from a to b. This integral does not depend on the path taken between the
two points.
4T€0",
4περΓ
EVALUATE Our result agrees with Eq. (23.14) and is correct for positive
or negative q.
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