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By integrating the electric field as in Eq. (23.17), find the potential at a distance r from a point charge q. 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. 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 dl is in the radial direction and has magnitude dr. Writing di = fdr, we have from Eq. (23.17) de= direction V - 0 = V = E; di 2 L rdr dr 4TEor = 0 4TEor, 4TEor, 4TEor

By integrating the electric field as in Eq. (23.17), find the potential at a distance r from a point charge q. 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. 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 dl is in the radial direction and has magnitude dr. Writing di = fdr, we have from Eq. (23.17) de= direction V - 0 = V = E; di 2 L rdr dr 4TEor = 0 4TEor, 4TEor, 4TEor

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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.
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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