(a) Explanation Given: Attractive force between two particles is given as F = k q 1 q 2 r 2 . Calculation: Let the polar rectangle makes an angle θ from the point P . The distance of the polar rectangle from the point P will be r 2 + d 2 . The diagram will look as shown below: Since area of polar rectangle is Δ r × Δ θ and charge density of the disk is ρ coulombs per square meter. Therefore, total charge on the polar rectangle will be ρ r ( Δ r × Δ θ ) . Vertical component of the force exerted by the polar rectangle on the point P can be written as: ⇒ F v e r t i c a l = F t o t a l sin ( θ ) Where F t o t a l = k q 1 q 2 r 2 + d 2 = k Q 1 ρ r ( Δ r × Δ θ ) r 2 + d 2 ∴ (b) To determine To explain: Why the force on the point P due to entire circular disk is given by the integral F = k Q ρ d ∫ 0 2 π ∫ 0 R r d r d θ ( r 2 + d 2 ) 3 2

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 16.5, Problem 59E

(a)

To determine

To prove:

That the vertical component of force on a point P due to a small polar rectangle of size $Δ r \times Δ θ$ located at distance r.

(b)

To determine

To explain:

Why the force on the point P due to entire circular disk is given by the integral

$F = k Q ρ d \int_{0}^{2 π} \int_{0}^{R} \frac{r d r d θ}{{(r^{2} + d^{2})}^{\frac{3}{2}}}$

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Chapter 16.5, Problem 58E

Chapter 16.5, Problem 60E

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Chapter 16 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 16.1 - Prob. 1PQ Ch. 16.1 - Prob. 2PQ Ch. 16.1 - Prob. 3PQ Ch. 16.1 - Prob. 4PQ Ch. 16.1 - Prob. 5PQ Ch. 16.1 - Prob. 6PQ Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E

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To prove: That the vertical component of force on a point P due to a small polar rectangle of size Δ r × Δ θ located at distance r .

Solution Summary: The author explains how the vertical component of force exerted by the polar rectangle on the point P can be written as follows:

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To prove:

That the vertical component of force on a point P due to a small polar rectangle of size $Δ r \times Δ θ$ located at distance r.

To explain:

Why the force on the point P due to entire circular disk is given by the integral

$F = k Q ρ d \int_{0}^{2 π} \int_{0}^{R} \frac{r d r d θ}{{(r^{2} + d^{2})}^{\frac{3}{2}}}$

Want to see the full answer?

Chapter 16 Solutions

To prove: That the vertical component of force on a point P due to a small polar rectangle of size Δ r × Δ θ located at distance r .

Solution Summary: The author explains how the vertical component of force exerted by the polar rectangle on the point P can be written as follows:

Concept explainers

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To prove: That the vertical component of force on a point P due to a small polar rectangle of size Δr×Δθ located at distance r.

To explain: Why the force on the point P due to entire circular disk is given by the integral F=kQρd∫02π∫0Rrdrdθ(r2+d2)32

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Chapter 16 Solutions

To prove:

That the vertical component of force on a point P due to a small polar rectangle of size $Δ r \times Δ θ$ located at distance r.

To explain:

Why the force on the point P due to entire circular disk is given by the integral

$F = k Q ρ d \int_{0}^{2 π} \int_{0}^{R} \frac{r d r d θ}{{(r^{2} + d^{2})}^{\frac{3}{2}}}$