Charged plate d R Ar FIGURE 21 59. According to Coulomb's Law, the attractive force between two electric charges of magnitude q1 and q2 separated by a distancer is kq192/r2 (k is a constant). Let F be the net force on a charged particle P of charge Q coulombs located d centimeters above the center of a circular disk of radius R, with a uniform charge distribution of density p coulombs per square meter (Figure 21). By symmetry, F acts in the vertical direction. (a) Let R be a small polar rectangle of size Ar x AÐ located at distance r. Show that R exerts a force on P whose vertical component is kpQd r Δr Δθ (r2 + d²)3/2 (b) Explain why F is equal to the following double integral, and evaluate: c27 r dr de F = kpQd (r2 + d²)3/2

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Charged plate
d
R
Ar
FIGURE 21
59.
According to Coulomb's Law, the attractive force between two
electric charges of magnitude q1 and q2 separated by a distancer is
kq192/r2 (k is a constant). Let F be the net force on a charged particle P
of charge Q coulombs located d centimeters above the center of a circular
disk of radius R, with a uniform charge distribution of density p coulombs
per square meter (Figure 21). By symmetry, F acts in the vertical direction.
(a) Let R be a small polar rectangle of size Ar x AÐ located at distance
r. Show that R exerts a force on P whose vertical component is
kpQd
r Δr Δθ
(r2 + d²)3/2
(b) Explain why F is equal to the following double integral, and evaluate:
c27
r dr de
F = kpQd
(r2 + d²)3/2
Transcribed Image Text:Charged plate d R Ar FIGURE 21 59. According to Coulomb's Law, the attractive force between two electric charges of magnitude q1 and q2 separated by a distancer is kq192/r2 (k is a constant). Let F be the net force on a charged particle P of charge Q coulombs located d centimeters above the center of a circular disk of radius R, with a uniform charge distribution of density p coulombs per square meter (Figure 21). By symmetry, F acts in the vertical direction. (a) Let R be a small polar rectangle of size Ar x AÐ located at distance r. Show that R exerts a force on P whose vertical component is kpQd r Δr Δθ (r2 + d²)3/2 (b) Explain why F is equal to the following double integral, and evaluate: c27 r dr de F = kpQd (r2 + d²)3/2
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