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Unformatted text preview: Electric Fields: Problem 1 (medium): Three charges, q 1 , q 2 , and q 3 are arranged at the vertices of the triangle shown in the figure below. Find the electric force on the charge q 3 . & & ¡ & ¢ & ¡ ¡ & & ¢ £ Solution: The distance between q 3 and q 1 is d 13 = l . The unit vector pointing from charge q 1 to charge q 3 is ˆ r 13 = cos θ ˆ i + sin θ ˆ j (1) Therefore, the force on q 3 due to charge q 1 is F 31 = q 3 q 1 4 π² d 2 13 ˆ r 13 = q 3 q 1 4 π² l 2 ‡ cos θ ˆ i + sin θ ˆ j · (2) The distance between q 3 and q 2 is d 23 = l . The unit vector pointing from charge q 2 to charge q 3 is ˆ r 23 = cos θ ˆ i + sin θ ˆ j (3) Therefore, the force on q 3 due to charge q 2 is 1 F 32 = q 3 q 2 4 π² d 2 23 ˆ r 23 = q 3 q 2 4 π² l 2 ‡ cos θ ˆ i + sin θ ˆ j · (4) Therefore, the total force on q 3 due to both charges is F 3 = F 31 + F 32 (5) The xcomponent of this force is F 3 x = F 31 ,x + F 32 ,x = q 3 q 1 4 π² l 2 cos θ q 3 q 2 4 π² l 2 cos θ = 0 (6) The ycomponent of this force is F 3 y = F 31 ,y + F 32 ,y = q 3 q 1 4 π² l 2 sin θ + q 3 q 2 4 π² l 2 sin θ = q 3 ( q 1 + q 2 ) 4 π² l 2 sin θ (7) Problem 2 (mediumhard): A rod has uniform linear charge density λ , and is bent into the shape of an arc of radius R . We define the origin to be the center of curvature of the arc. As shown in the figure, the lower end of the arc is displaced from the origin at an angle θ i to the xaxis, while the upper end of the arc is displaced from the origin at an angle θ f to the xaxis. Find the electric field at the origin due to the arc. 2 & ¡ ¢ & £ & ¡ ¤ ¥ Solution: The rod is an example of a continuous distribution of charge. We first seek to find the electric field d E due to an element of this arc. Such an element has a charge dq = λRdθ , and is located at some angle θ to the origin, at a distance R from the origin. The displacement vector from this element of charge to the origin is: r θ = R ‡ cos θ ˆ i sin θ ˆ j · ˆ r θ = cos θ ˆ i sin θ ˆ j (1) Therefore, the electric field at the origin contributed by this element of the arc is d E = dq 4 π² R 2 ˆ r θ = λRdθ 4 π² R 2 ‡ cos θ ˆ i sin θ ˆ j · = λdθ 4 π² R ‡ cos θ ˆ i sin θ ˆ j · (2) The total electric field at the origin due to the entire arc is obtained by integrating over these elements of charge: 3 E = Z d E = Z θ f θ i λdθ 4 π² R ‡ cos θ ˆ i sin θ ˆ j · = λ 4 π² R Z θ f θ i dθ ‡ cos θ ˆ i sin θ ˆ j · = λ 4 π² R ‡ sin θ ˆ i + cos θ ˆ j · fl fl fl fl fl θ f θ i = E x ˆ i + E y ˆ j (3) where E x = λ 4 π² R (sin θ f + sin θ i ) E y = λ 4 π² R (cos θ f cos θ i ) (4) Problem 3 (mediumhard): A spherical insulating shell has inner radius a and outer radius b , and is filled with uniform charge density ρ . Surrounding this shell, and concentric with it, is a spherical conducting shell of inner radius b and outer radius c . The conducting shell carries a net charge...
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 Spring '09
 WAUNG
 Electric charge

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