Solution MC Hwk (5)

Solution for Multiple-Choice Homework 5 Gauss’ Law Solution to Multiple-Choice Homework Problem 5.1(Electric Field for Spherical Shells) Problem: The figure to the right shows two charged spherical shells. The inner shell has radius a and charge density σa = −2σ. The outer shell has radius b and charge density σb = +3σ. Calculate the electric field in Region II between the two shells (a < r < b).

y III II

Select One of the Following:

a I Air

(a) 0 2a2 σ rˆ (b-Answer) − ε0 r 2 a2 σ (c) + rˆ 2πε0 r2 5πa2 σ (d) − rˆ 4πε0 r2 πa2 σ (e) + rˆ 4πε0 r2 −2πa2 σ + 3πb2 σ (f) rˆ 4πε0 r2

Air

x b

Air

Solution The charge enclosed by a Gaussian surface in Region II is the charge of the inner shell, Qenc = −8πa2 σ.The electric field between the two shells is then 2 2 ~ II = Qenc = (4πa )(−2σ) rˆ = −2a σ rˆ E 4πε0 r2 4πε0 r2 ε0 r 2

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 5.2(Flux Through Hoop Normal Perpendicular) Problem: A square hoop with sides of length 3.0m is in a uniform electric field with magnitude 1.0 × 105 N C . Its normal is perpendicular to the field (careful here). What is the flux through the hoop? Select One of the Following: (a-Answer) 0 2 (b) 3.0 × 105 N Cm 2 (c) 1.0 × 105 N Cm 2 (d) 6.0 × 105 N Cm 2 (e) 9.0 × 105 N Cm

Solution The electric flux through the loop is 0, because the electric field is perpendicular to the surface normal, so no field lines pass through the surface. This could be a loop in the x − y plane with an electric field in the x direction. The normal for the loop is in the z direction. Since the normal and the field are perpendicular, the flux must be zero. Total Points for Problem: 3 Points

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Solution to Multiple-Choice Homework Problem 5.3(Flux Through the Surface for Different Charge Locations) Problem: The figure to the right shows a spherical Gaussian surface. A point charge +Q is placed at one of three points inside the surface. Rank the electric flux out of the surface, φA , φB , and φC , for each placement.

C

Select One of the Following: (a-Answer) φA = φB = φC

A

(b) φA > φB > φC

B

(c) φA < φB < φC (d) φA = φB < φC (e) φA < φB = φC (f) φA > φB = φC

Solution The flux is the same for each scenario listed. The electric flux depends only on the charge inside the surface, not . the location. Gauss’ Law tells us φ = Qǫenc 0 Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 5.4(When is Gauss’ Law True?) Problem: For what kind of charge distributions does Gauss’s law hold? Select One of the Following: (a) only highly symmetric charge distributions (b) any symmetric charge distribution (c-Answer) any charge distribution (d) positive charge distributions (e) negative charge distributions Solution Total Points for Problem: 3 Points Gauss’ Law is always true.

Solution to Multiple-Choice Homework Problem 5.5(Flux Through Wedding Ring Normal Parallel) Problem: A wedding ring is a circle with radius 1.0cm. If the normal to the surface bounded by the ring is parallel to the earth’s electric field of 150 N C , what is the electric flux through the ring? Select One of the Following: 2 (a) 1.5 × 10−2 N Cm 2 (b) 2.3 × 10−2 N Cm 2 (c-Answer) 4.7 × 10−2 N Cm

2

2 (d) 3.5 × 10−1 N Cm

(e) 0 Solution Since the field is uniform and the normal is parallel to the field, the electric flux through the loop is ¶ µ N N 2 π(1 × 10−2 m)2 = 4.7 × 10−2 m2 φ = EA = EπR = 150 C C since the normal is parallel to the field. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 5.6(Electric Flux of Sphere Charged by Battery) Problem: If a metal sphere of radius 1cm is charged with a “D” cell battery, a charge of 1.7 × 10−12 C is developed. If this sphere is then placed in a cubic box with edges of length 30.0cm. What is the electric flux out of the box? Select One of the Following: (a) 2.7 × 10−15 Nm2 /C (b) 4.6 × 10−14 Nm2 /C (c) 1.7 × 10−4 Nm2 /C (d) 5.2 × 10−3 Nm2 /C (e-Answer) 0.19Nm2 /C Solution By Gauss’ Law, the magnitude of the total electric flux out of a cube containing a net charge Q is proportional to the charge Nm2 1.7 × 10−12 C Qenclosed = 0.19 = φ= 2 C ǫ0 C 8.85 × 10−12 Nm2 Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 5.7(Charge to Produce Tabletop Accelerator Fields) Problem: A tabletop accelerator uses lasers to accelerate electrons. The accelerator produces an electric field 11 N of 100, 000, 000, 000 N C = 1 × 10 C . How much charge would I have to put on a pith ball of radius 0.5cm to 11 N produce a field of 1 × 10 C at its surface? Select One of the Following: (a) 6 × 10−2 C (b-Answer) 3 × 10−4 C (c) 5 × 102 C (d) 4 × 106 C (e) 2 × 10−2 C Solution

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Outside a spherical system, the field is that of a point charge at the center of the system. If R is the radius of the pith ball and Q its charge, then the field at its surface is E = kQ/R2 . The charge required for the given field is then −3 (1 × 1011 N m)2 ER2 C )(5 × 10 Q= = = 3 × 10−4 C 2 Nm k 8.99 × 109 C2 Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 5.8(Electric Field for Spherical Shells) Problem: The figure to the right shows two charged spherical shells. The inner shell has radius a and charge density σa = +2σ. The outer shell has radius b and charge density σb = −3σ. Calculate the electric field at points in Region I inside the inner shell, at a radius of r < a.

y III

Select One of the Following:

II a I Air

(a-Answer) 0 σ (b) − rˆ 4πε0 r2 σ rˆ (c) + 4πε0 r2 4πa2 σ (d) − rˆ 4πε0 r2 4πa2 σ (e) + rˆ 4πε0 r2 4πa2 σ − 9πb2 σ (f) rˆ 4πε0 r2

Air

x b

Air

Solution A spherical Gaussian surface in Region I encloses no charge, therefore the field in Region I is zero. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 5.9(Charge in Gauss’ Law ) Problem: The figure to the right shows a surface and two point charges QA and QB . If Gauss’ law is applied to the surface, which of the charges must be included in the charge used in Gauss’ law? Select One of the Following: (a) neither charge should be included (b) both charges should be included

QA

(c-Answer) only charge QA should be included (d) only charge QB should be included

Solution Only the charge QA needs to be used in Gauss’ law. Gauss’ law applies to the charge inside the surface. 4

QB

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 5.10(Qualitative Direction of Field Lines In Region ) Problem: The two thin concentric spherical shells each have uniform surface charge densities. The total charge on the inner shell is +Q and the total charge on the outer shell is −Q. Which of the following best describes the direction of the electric field between the shells? Select One of the Following: (a-Answer) The electric field between the shells points radially outward from the center of the system. (b) The electric field between the shells points radially inward toward the center of the system. (c) The electric field between the shells points radially outward from the center of the system near the inner shell and changes direction to radially inward near the outer shell. (d) The electric field between the shells points radially inward from the center of the system near the inner shell and changes direction to radially outward near the outer shell. (e) The electric field lines are concentric circles around the inner shell. Solution The charge of the inner shell is positive, so the field lines point radially inward, since Gauss’ law applies only to the charge inside. Total Points for Problem: 3 Points

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