Homework-5-sln - HW5 PDF

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Solution for Open-Response Homework 5 Gauss’ Law Solution to Open-Response Homework Problem 5.1(Two shell Gauss’ Law Problem) Problem: A point charge with charge q is surrounded by two thin shells of radius a and b which have surface charge density σ a and σ b . (a)Specialize Gauss’ Law from its general form to a form appropriate for spherical symmetry. (b)Compute the electric field in region I. (c)Compute the electric field in region II. (d)Compute the electric field in region III.

Solution to Part(a) Gauss’ Law states that the electric flux, φ, out of a closed surface is φ=

Qenclosed ε0

where Qenclosed is the total charge enclosed in the surface. Flux is the area of a surface multiplied by the electric field in direction of the normal of the surface. For a spherically symmetric system, the electric field is radial and always points in the direction of the normal (or exactly opposite the direction)of a spherical Gaussian surface rˆ. The electric flux out of a spherical Gaussian surface of radius r is therefore, φ = E(r)A where A is the area of the Gaussian surface or using the surface area of a sphere φ = 4πr2 E(r). Applying Gauss’ Law to the flux yields φ = 4πr2 E(r) = or E(r) =

Qenclosed ε0

Qenclosed 4πε0 r2

Grading Key: Solution to Part (a) 5 Points Solution to Part(b) The total charge enclosed by a Gaussian surface placed in region I is Qenclosed = q, therefore the electric field is ~EI =

q rˆ 4πε0 r2

Grading Key: Solution to Part (b) 3 Points Solution to Part(c) The total charge of the spherical shell is 4πa2 σ a . The total charge enclosed by a Gaussian surface placed in region II is Qenclosed = q + 4πa2 σ a therefore the electric field in region II is 2 ~ II = q + 4πa σ a rˆ E 4πε0 r2

Grading Key: Solution to Part (c) 3 Points

1

Solution to Part(d) The total charge of the spherical shell is 4πb2 σ b . The total charge enclosed by a Gaussian surface placed in region III is Qenclosed = q + 4πa2 σ a + 4πb2 σ b therefore the electric field in region III is 2 2 ~ III = q + 4πa σ a + 4πb σ brˆ E 2 4πε0 r

Grading Key: Solution to Part (d) 3 Points Total Points for Problem: 14 Points

Solution to Open-Response Homework Problem 5.2(Computing Field at Finite Distance From Sphere) Problem: A sphere of radius a = 60cm carries a uniform volume charge density, ρ = +5×10−6 C/m3 and is centered on the origin. A larger spherical shell of radius b = 1.2m is concentric with the first and carries a uniform surface charge density, σ = −1.5 × 10−6 C/m2 .

Air Air

(a)What is the total charge of the sphere and the shell? (b)Draw the electric field map. Assign lines per Q as best you can. (c)Use Gauss’ law to find ~E at (0.5m, 0, 0). Compute both the symbolic and numeric forms of the field. ~ at (−2m, 3m, 0). Compute (d)Use Gauss’ law to find E both the symbolic and numeric forms of the field.

b

volume charge a I II III

Definitions ~ i ≡ Electric Field in Region I E

rvol = 0.6m ≡ Radius of Volume Charge

rshell = 1.2m ≡ Radius of Shell of Charge

ρ = 5 × 10−6 C/m3 ≡ Volume Charge Density of Sphere

σ = −1.5µC/m2 ≡ Surface Charge Density of Shell

r rvol

rshell

Solution to Part (a) The total charge of the volume charge is 4 C 4 Qvol = ( πr3vol )ρ = ( π)(0.6m)3 (5 × 10−6 3 ) = 4.5 × 10−6 C. 3 3 m

2

The total charge of the shell is 2 Qshell = 4πrshell σ = 4π(1.2m)2 (−1.5 × 10−6

C ) m2

= −2.7 × 10−5 C. Grading Key: Solution to Part (a) 4 Points Solution to Part(b) The previous step computes the charge of each object. Now, assign a number of field lines per charge. Let 3 lines = 4.5 × 10−6 C ⇒ 18 lines = 2.7 × 10−5 C. In region II, the total charge enclosed is Qvol = 3 lines outward. In region III, the total charge enclosed is ¶ µ 3lines = 15lines inward Qenc = Qshell + Qvol = −2.25 × 10−5 C = −2.25 × 10−5 C 4.5 × 10−6 C using Gauss’ Law (Version 0) to draw the field map. Grading Key: Solution to Part (b) 3 Points 1 point(s) (3 times) : Each correct region. Solution to Part (c) (a) Specialize Gauss’ law for Spherical Symmetry: Because of the spherical symmetry of the charge ~ r and Gauss’ Law applied to a spherical Gaussian surface distribution, the electric field can be written E = E(r)ˆ becomes, Z ~ ·n φe = (E ˆ )dA S

Substitute radial field and use rˆ · rˆ = 1,

φe =

Z

S

(E (r)ˆ r) · rˆ)dA

φe = E(r)

Z

dA

S

The integral is just the surface area of the Gaussian surface, Z dA = 4πr2 S

Apply Gauss’ Law φe = 4πr2 E(r) = or

Qenclosed ε0

~ = Qenclosed rˆ. E 4πǫ0 r2

This was not required for the problem, but I like to start at the beginning. (b) Determine the Region Containing the Field Point: We are asked to calculate the field at a point P at ~rP = (0.5m, 0). This point is a distance rP = 0.5m from the origin, so rP < rvol and the field point is in the volume charge in region I (c) Compute the Field in Region I: A Gaussian surface of radius r in region I (r < rvol ) encloses PART of the volume charge. The charge enclosed in the Gaussian surface is Qenc = 4 3 πr3 ρ, the charge density multiplied by the volume of the Gaussian surface. Applying the form of Gauss’ law for spherical symmetry derived earlier yields the electric field in region I, 4 πr3 ρ ~EI = Qenclosed rˆ = 3 rˆ 2 4πǫ0 r 4πǫ0 r2 or simplifying ~ I = ρr rˆ E 3ε0 3

~ at 0.5m: The displacement vector ~rP and its length is calculated above, the unit vector is (d) Compute E rˆP = x ˆ by observation.The electric field is −6 3 ~E(0.5m) = ρr ˆr = (+5 × 10 C/m )(0.5m) x ˆ C2 ) 3ε0 3(8.85 × 10−12 Nm 2

N ˆ, = 9.4 × 104 x C Grading Key: Solution to Part (c) 4 Points Solution to Part (d) p √ ~ at (−2m, 3m): The point ~rP = (−2m, 3m, 0) is a distance rP = (2m)2 + (3m)2 = 13m from Compute E √ the origin.This distance is greater than the radius of the shell, 13m = 3.6m > 1.2m, so the field point is outside of all charge in region III.The unit vector of ~rP is (−2m, 3m, 0) ~rP √ = . rP 13m µ ¶ −2 3 rˆ = √ , √ , 0 13 13 A spherical Gaussian surface in region III always encloses all of the volume charge and the shell of charge, so the charge enclosed by a Gaussian surface in region III is Qenc = Qvol + Qshell . Substitute the charge enclosed into Gauss’ law for spherical symmetry, rˆP =

~ III = Qenclosed rˆ = Qvol + Qshell rˆ E 4πǫ0 r2 4πǫ0 r2 Substitute rP and rˆP into Gauss’ law 6 −5 C) ~ = Qvol + Qshellrˆ = (4.5 × 10 C − 2.7 × 10 E √ 2 2 C 4πǫ0 r 4π (8.85 × 10−12 Nm2 )( 13m)2 µ ¶ N 2 −3 √ , √ , 0 . = 1.6 × 104 C 13 13

Grading Key: Solution to Part (c) 6 Points Total Points for Problem: 17 Points

4

µ

3 −2 √ , √ , 0 13 13

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