15 - 15.RQ.20 The uniform spherical shell of charge on the...

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15.RQ.20 The uniform spherical shell of charge on the surface of the ball contributes zero electric field inside the surface. Since the polarization of the molecules is proportional to the applied electric field, none of the molecules are polarized.
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15.RQ.22 A spark will be triggered when the maximum electric field exceeds about 3 × 10 6 N/C. The largest field is between the plates. So a spark will occur when Q / A ε 0 3 × 10 6 N/C Q ≈ε 0 A (3 × 10 6 N/C) (9 × 10 -12 C 2 N m 2 )( π )(2.5m) 2 × 10 6 N/C) 5.3 × 10 -4 C
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15.RQ.25 const. Electric field of a charged disk, very near the center of the disk. 1/ r Electric field of a rod at a location perpendicular to the rod, where r << length of rod. 1/ r 2 Electric field of a point charge or of a uniformly charged sphere (outside the sphere). 1/ r 3 Electric field of a dipole, far from the dipole. 1/ r 5 Force on an external charge due to an induced dipole.
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15.P.26 Charged tape (a) x y d θ .02 m .04 m Δ E x Δ E 1. Cut up distribution into pieces and draw Δ E. Problem statement specifies 3 pieces, each 2 cm wide by 4 cm high. Δ E for top piece is shown in diagram. It is clear from the diagram that we only need the x component, since the y components will cancel. 2. Algebraic expression for Δ E for each piece: Consider each piece as a point charge, located at center of rectangle. Then: Δ E = 1 4 πε 0 Δ Q d 2 , and we want x - component : Δ E x = 1 4 πε 0 Δ Q d 2 cos θ d = x 2 + y 2 , and cos θ= x d = x x 2 + y 2 , and Δ Q = Q A () Δ A (where Q = total charge on tape, A = total area, Δ A= Δ y Δ z = area of small piece.) Δ E x = 1 4 πε 0 Q A Δ A x 2 + y 2 x x 2 + y 2 = 1 4 πε 0 Q A Δ A x x 2 + y 2 3 2 4. Add up all contributions: Since we are at point in middle, y and z components of E will add up to zero, and E net will be in x direction. We only need to find x-component of Δ E for each piece. E x E x = 1 4 πε 0 Q A ( ) Δ A x i x i 2 + y i 2 3 2 i = 1 3 For this calculation, 1 4 πε 0 Q A Δ x Δ y = 9 × 10 9 N m 2 C 2 4 × 10 8 C .02m .12m .02m .04m = 120 N m 2 C Only the last term depends on distance, so we'll calculate it for each piece: Piece x y z x/(x 2 +y 2 ) 1.5 1 .03 m .04 m 0 240 2 .03m 0.00 m 0 1111 3 .03 m .04 m 0 240 Total: 1591 So E x = 120 N m 2 C 1591 m -2 ( ) = 1.91 × 10 5 N C 4. Check that result is reasonable: (a) The units come out right (N/C). (b) This is less than the critical field for breakdown of air (3x10 6 N/C), so it is a possible value. (c) Also, a point charge of 4x10 -8 C located 3 cm away would make a field of (9x10 9 )(4x10- 8 )/(.01) 2 = 4x10 5 N/C, and our E is smaller than that, as it should be, since almost all of the tape is farther away than 3 cm. .
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(b) x y z d θ .03 m .01 m Δ E x Δ E 1. Cut up distribution into pieces and draw Δ E. Problem statement specifies 8 pieces, each 1 cm wide by 3 cm high. Δ E is shown.
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This note was uploaded on 05/24/2011 for the course PCS 228 taught by Professor Joseph during the Spring '10 term at Ryerson.

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15 - 15.RQ.20 The uniform spherical shell of charge on the...

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