Explanation Given: The plate is r 2 = ( 1 6 sin 2 θ + 1 9 cos 2 θ ) − 1 with the disk x 2 + y 2 ≤ 1 that is r = 1 removed and the charge density is ρ ( r , θ ) = 3 r − 4 Formula used: The total charge of a region D with a given charge density δ ( x , y ) is the solution of the integration ∬ D ρ ( r , θ ) r d r d θ (in 2-dimention). Calculation: Here the region is the given plate, so, 1 ≤ r ≤ ( 1 6 sin 2 θ + 1 9 cos 2 θ ) − 1 / 2 ; 0 ≤ θ ≤ 2 π Then ∬ D ρ ( r , θ ) r d r d θ = ∫ 0 2 π ∫ 1 ( 1 6 sin 2 θ + 1 9 cos 2 θ ) − 1 / 2 3 r − 4 r d r d θ = 3 ∫ 0 2 π ∫ 1 ( 1 6 sin

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.5, Problem 10E

To determine

The total charge on a plate assuming a charge density $ρ (r, θ)$

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Chapter 16.5, Problem 9E

Chapter 16.5, Problem 11E

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Chapter 16 Solutions

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The total charge on a plate assuming a charge density ρ ( r , θ )

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The total charge on a plate assuming a charge density ρ ( r , θ )

Solution Summary: The author explains how to determine the total charge on a plate assuming the charge density ( r.

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The total charge on a plate assuming a charge density ρ(r,θ)

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The total charge on a plate assuming a charge density $ρ (r, θ)$