PRACTICE: Consider the setup shown in the figure below, where the arc is a semicircle with radius r. The total charge Q is negative, and distributed uniformly on the semicircle. The charge on a small segment with angle A0 is labeled Aq. The magnitude of the x-component of the electric field at the center, due to Aq, is given by !! II III IV 1. AE, = k|Sq| (cos@) r よ 6. AE, = 2. AE, = kSyi sin e 7. AE, = kSy|l (sin@) r 3. AE, = 1:|A4| (cos@) r 8. ΔΕ, = 4. AE, = k|Aq! (sin@) r kAq| cos e 9. JE, = 5.JE, = 10. JE, = k|yr If correct plese explein,

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PRACTICE: Consider the setup shown in the figure below, where the arc is a semicircle with
radius r. The total charge Q is negative, and distributed uniformly on the semicircle. The
charge on a small segment with angle Ae is labeled Aq. The magnitude of the x-component of
the electric field at the center, due to Aq, is given by
!!
II
III IV
1. AE, = k|Aq| (cos6)r
6. AE, =
kSyi sin e
2. AE, =
7. AE; = k|A| (sin@) r
3. AE, = 1:|Ayl (cos @) r
k: JAg| sin e
8. AE, =
4. AE, = kAq} (sin #) r
kAy cos e
9. JE, =
5.AE, = A cos #
10. AE, = k|Aylr
%3D
If correct plese expleis.
Transcribed Image Text:PRACTICE: Consider the setup shown in the figure below, where the arc is a semicircle with radius r. The total charge Q is negative, and distributed uniformly on the semicircle. The charge on a small segment with angle Ae is labeled Aq. The magnitude of the x-component of the electric field at the center, due to Aq, is given by !! II III IV 1. AE, = k|Aq| (cos6)r 6. AE, = kSyi sin e 2. AE, = 7. AE; = k|A| (sin@) r 3. AE, = 1:|Ayl (cos @) r k: JAg| sin e 8. AE, = 4. AE, = kAq} (sin #) r kAy cos e 9. JE, = 5.AE, = A cos # 10. AE, = k|Aylr %3D If correct plese expleis.
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