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PHYS2102 (Aktas) Practice Test 2 Hill, Ryan - [email protected]

Problem 1: Point charges located at 3, 8, and 11 cm along the x - axis are

represented in the figure.

Randomized Variables

q = 1.25 μC

Part (a) What is the force on the charge located at x = 8 cm given that q = 1.25 μC in N?

Correct Answer Student AnswerFinal

Grade

F = 20 F = 19.97 96%

Grade Detail

Correct Student Feedback

Final Answer Credit 100%

Submission Detail

Answer Hints Feedback Totals

1 F = 0.37 4% 0% 0% 4%

2 F = 19.97 0% 0% 0% 0%

Totals 4% 0% 0% 4%

Problem 2: A 3-D printer lays down a semicircular arc of positively charged

plastic with a radius R = 3.5 cm, and a linear charge density of λ = +2 μC/m. After

the printer has finished the arc, the stylus moves to the center of the arc as shown.

The minute segment of the plastic arc highlighted in the diagram subtends an angle

dθ. Note the measurement of the angle θ shown in the figure.


R = 3.5 cm

λ = 2 μC/m

Part (a) Input a symbolic expression for the charge dq on the segment of charge of size dθ in terms of given parameters.

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https://www.theexpertta.com/Common/GradeSheetClassAssignments.aspx?eid=453

https://www.theexpertta.com/Common/GradeSheetClassAssignmentProblems.aspx?eid=453&aid=2293


Grade

dq = λ R dθ dq = λ R dθ 100%

Grade Detail



Submission Detail


1 dq = λ R dθ 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (b) Input a symbolic expression for the magnitude of the electric force, exerted by the minute segment of plastic subtending the arc dθ on an

electron (charge e), within the stylus at the center of the arc. Express your answer in terms of given parameters and fundamental constants.


Grade

dF = ( k e λ dθ )/R dF = ( e λ k dθ )/( R ) 100%

Grade Detail



Submission Detail


1 dF = ( e λ k dθ )/( R ) 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (c) Find this indefinite integral, but do not evaluate the limits, from Part (b) to determine the magnitude of the force exerted on the electron in the

stylus tip by the entire line of charged plastic.


Grade

F = ( k e λ sin(θ) )/R F = ( e λ k/R ) ( sin(θ) ) 100%

Grade Detail



Submission Detail


1 F = ( e λ k/R ) ( sin(θ) ) 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (d) Select the limits of integration that would result in the correct calculation of the force exerted by the entire line of plastic on an electron in the

stylus as shown in the diagram.


Grade

θ = –π/2 to θ = π/2 θ = –π/2 to θ = π/2 100%

Grade Detail



Submission Detail


1 θ = –π/2 to θ = π/2 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (e) Using the limits from Part (d) and the expression from Part (c) to determine an expression for the magnitude of the electrical force exerted by

the positively charged line of plastic on the electron in the stylus.


Grade

F = ( 2 k e λ )/R F = ( 2 e λ k )/( R ) 100%

Grade Detail



Submission Detail


1 F = ( 2 e λ k )/( R ) 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (f) Calculate the magnitude of the electric force (in Newtons) exerted by the positively charged line of plastic on the electron in the stylus.


Grade

F = 1.644E-13 F = 1.64 * 10-13

F = 1.64E-13

100%

Grade Detail



Submission Detail


1 F = 1.64 * 10-13

F = 1.64E-13

0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (g) What direction is the electric force exerted on the electron in the stylus?


Grade

Left Left 100%

Grade Detail



Submission Detail


1 Left 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Problem 3: A positive charge of magnitude Q1 = 3.5 nC is located at the origin. A

negative charge Q2 = -9.5 nC is located on the positive x-axis at x = 3.5 cm from the

origin. The point P is located y = 4.5 cm above charge Q2.


Q1 = 3.5 nC

Q2 = -9.5 nC

x = 3.5 cm

y = 4.5 cm

Part (a) Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.


Grade

100%

Ex,1 = 5944 Ex,1 = 5945.9

Ex,1 = 5946

Grade Detail



Submission Detail


1 Ex,1 = 5945.9

Ex,1 = 5946

0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (b) Calculate the y-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.


Grade

Ey,1 = 7642 Ey,1 = 7653.65

Ey,1 = 7654

100%

Grade Detail



Submission Detail


1 Ey,1 = 7653.65

Ey,1 = 7654

0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (c) Calculate the y-component of the electric field at point P due to the Charge Q2. Write your answer in units of N/C.


Grade

Ey,2 = -42180 Ey,2 = - 42279.7

Ey,2 = -42280

91%

Grade Detail



Submission Detail


1 Ey,2 = 42279.7

Ey,2 = 42280

4% 0% The answer provided was not correct. We have recognized the following,

- Your answer appears to be off by a factor of -1. Your answer contains a

sign error.

5% 9%

2 Ey,2 =

- 42279.7

Ey,2 = -42280

0% 0% 0% 0%

Totals 4% 0% 5% 9%

Part (d) Calculate the y-component of the electric field at point P due to both charges. Write your answer in units of N/C.


Grade

Ey,2 = -34530 Ey,2 = - 34637.7

Ey,2 = -34640

100%

Grade Detail



Submission Detail


1 Ey,2 = - 34637.7

Ey,2 = -34640

0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (e) Calculate the magnitude of the electric field at point P due to both charges. Write your answer in units of N/C.


Grade

E = 35040 E = 35037.8

E = 35040

96%

Grade Detail



Submission Detail


1 E = 595547.1

E = 595500

4% 0% 0% 4%

2 E = 35037.8

E = 35040

0% 0% 0% 0%

Totals 4% 0% 0% 4%

Part (f) Calculate the angle in degrees of the electric field at point P relative to the positive x-axis.


Grade

θ = -80.23 θ = 8.005 77%

Grade Detail


see

above

The answer provided was not correct. We have recognized the following,

- Your answer appears to be off by a factor of -1/10. Ensure you have represented the number in the correct

units. Your answer contains a sign error.

15%


Submission Detail


1 θ =

52.1

4% 0% 0% 4%

2 θ =

81.99

4% 0% Note: Feedback provided here but not accessed during assignment.


- Your answer appears to be off by a factor of -1. Your answer contains a sign error.

0% 4%

3 θ =

8.005

0% 0% Note: Feedback provided here but not accessed during assignment.


- Your answer appears to be off by a factor of -1/10. Ensure you have represented

the number in the correct units. Your answer contains a sign error.

0% 0%

Totals 8% 0% 0% 8%

Problem 4: Point charges of 0.21 μC and 0.45 μC are placed 0.3 m apart.


q1 = 0.21 μC

q2 = 0.45 μC

d = 0.3 m

Part (a) At what point along the line between them is the electric field zero? Give your answer in m from 0.21 μC charge.


Grade

x = 0.1218 x = 0.12 100%

Grade Detail



Submission Detail


1 x = 0.12 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (b) What is the magnitude of the electric field halfway between them in N/C?

Correct Answer Student AnswerFinalGrade

E = 96000 E = 95894

E = 95890

92%

Grade Detail



Submission Detail


1 E = 83906.6E = 83910

4% 0% 0% 4%

2 E = 263706.6E = 263700

4% 0% 0% 4%

3 E = 95894

E = 95890

0% 0% 0% 0%

Totals 8% 0% 0% 8%

Problem 5: A 5.6 g charged insulating ball hangs on a 1.15 cm long string in a

uniform horizontal electric field as shown in the figure.


m = 5.6 g

l = 1.15 cm

Part (a) Given the charge on the ball is 33 μC , find the strength of the field in N/C.


E = 233.6 E = 234.72

E = 234.7

96%

Grade Detail



Submission Detail


1 E =234487.27

E =234500

4% 0% Note: Feedback provided here but not accessed during assignment.The answer provided was not correct. We have recognized the following,

- Your answer appears to be off by a factor of 1000. Ensure you haverepresented the number in the correct units.

0% 4%

2 E =234.72

E = 234.7

0% 0% 0% 0%

Totals 4% 0% 0% 4%

Problem 6: A plastic rod of length d = 2 m lies along the x-axis with its midpoint atthe origin. The rod carries a uniform charge density λ = 8 nC/m. The point P is located

along the y-axis at a distance y = 75 cm above the origin. The z-direction comes outof the page.


d = 2 mλ = 8 nC/m

y = 75 cm

Part (a) At point P, by symmetry there will be no electric field in the _____________ direction


Grade

x and z directions x and z directions 100%

Grade Detail



Submission Detail


1 x and z directions 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (b) Write an equation for the y-component of the electric field at P due to a thin slice of the rod of thickness dx located at point x.


100%

Grade Detail



Submission Detail


1 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (c) Integrate the electric field contributions over the length of the rod to find an equation for the electric field at point P.


Grade

100%

Grade Detail



Submission Detail


1 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (d) Calculate the magnitude of the electric field at point P in units of N/C.


E = 153.4 E = 153.43E = 153.4

96%

Grade Detail



Submission Detail


1 E = 1912.5

E = 1913

4% 0% 0% 4%

2 E = 153.43E = 153.4

0% 0% 0% 0%

Totals 4% 0% 0% 4%

Problem 7: A uniformly charged rod of length L = 2.1 m lies along the x-axis with its

right end at the origin. The rod has a total charge of Q = 8 μC. A point P is located onthe x-axis a distance a = 2 m to the right of the origin.


L = 2.1 mQ = 8 μC

a = 2 m

Part (a) Consider a thin slice of the rod of thickness dx located a distance x away from the origin. What is the direction of the electric field at point Pdue to the charge on this thin slice of the rod?


Grade

Positive x-direction Positive x-direction 100%

Grade Detail



Submission Detail


1 Positive x-direction 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (b) Write an equation for the electric field at point P due to the thin slide of the rod dE. Give your answers in terms of the variables Q, L, x, a, dx,and the Coulomb constant, k.

Final

Correct Answer Student Answer Grade

dE = k Q dx/( L ( a + x )2 ) dE = ( Q k dx )/( L ( a + x )2 ) 83%

Grade Detail



Submission Detail


1 dE = ( Q k )/( a ( L + a ) ) 4% -Since the charge Q is uniformly distributed overthe rod, what portion of the total charge will be

located on the slice of thickness dx?

3% 0% 7%

2 dE =( ( Q k )/L ) ( ( 1/a ) - ( 1/( L + a ) ) )

4% -Since the slice is thin, it can be treated as a pointcharge. How far is the point P away from this

charge. Treat x as a positive number.-How is the electric field determined from the

charge and distance between the point and thecharge?

6% 0% 10%

3 dE = ( Q k dx )/( L ( a + x )2 ) 0% 0% 0% 0%

Totals 8% 9% 0% 17%

Part (c) Integrate the electric field contributions from each slice over the length of the rod to write an equation for the net electric field E at point P.


Grade

E = k Q/( a ( a + L ) ) E = ( Q k )/( a ( L + a ) ) 100%

Grade Detail



Submission Detail


1 E = ( Q k )/( a ( L + a ) ) 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (d) Calculate the magnitude of the electric field E in kilonewtons per coulomb (kN/C) at point P due to the charged rod.


E = 8.771 E = 8.77 96%

Grade Detail



Submission Detail


1 E = 8.7 * 109

E = 8.7E+9

4% 0% 0% 4%

2 E = 8.77 0% 0% 0% 0%

Totals 4% 0% 0% 4%

Problem 8: A thin rod of length L has a positive charge Q which is uniformly

distributed across the rod. The rod lies on the x-axis. Point P is in front of the rod, onthe x-axis, a distance d from the rod.

Part (a) Write an expression for the linear charge density λ of the rod.


λ = Q/L λ = Q/L 100%

Grade Detail



Submission Detail


1 λ = Q/L 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (b) Find the y-component of the electric field, Ey, at point P.


Grade

Ey = 0 Ey = 0 100%

Grade Detail



Submission Detail


1 Ey = 0 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (c) Consider an infinitesimal portion dQ of the rod. Write an expression for the magnitude dEx of the x-component of the electric field, created by

this portion, at a point a distance r away, from P.


Grade

75%

Grade Detail



Submission Detail


1 25% 0% Note: Feedback provided here but not accessed during assignment.The distance from the charge element is r.

0% 25%

2 0% 0% 0% 0%

Totals 25% 0% 0% 25%

Part (d) Express the infinitesimal charge portion dQ through the linear charge density and the length of the segment, dr.


dQ = λ dr dQ = λ dr 100%

Grade Detail



Submission Detail


1 dQ = λ dr 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (e) Write an integral that is an expression for the x-component Ex of the electric field at point P.


Grade

64%

Grade Detail



Submission Detail


1 33% -Integration is the summationof all the dEx on the rod.

3% Note: Feedback provided here but notaccessed during assignment.The closest portion of charge is a distance d from

point P, the farthest a distance L+d.

0% 36%

2 0% 0% 0% 0%

Totals 33% 3% 0% 36%

Part (f) Calculate an expression for the x-component Ex of the electric field at point P, in terms of Q, L, d, and k.


Grade

100%

Grade Detail



Submission Detail


1 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (g) Find an expression for the x-component Ex of the electric field at a large distance d from the rod, d >> L. Since d >> L you may assume that

L/d ≈ 0.


Ex = k Q/d2 Ex = ( ( k Q )/d2 ) 100%

Grade Detail



Submission Detail


1 Ex = ( ( k Q )/d2 ) 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Problem 9: Two charges lie in a line along the x axis. Charge 1 is q1 = 0.9 C and

charge 2 is q2 = 2.2 C. They are each a distance of d = 0.038 m from the origin.


q1 = 0.9 C

q2 = 2.2 C

d = 0.038 m

Part (a) What is the distance on the x-axis from the origin at which the electric field will be zero. Give your answer in meters.


xo = 8.353E-3 xo = 0.0084 89%

Grade Detail



Submission Detail


1 xo =

0.0463

4% 0% 0% 4%

2 xo =

0.0159

4% -Since both charges are positive the zero field will be between them. ( -d <x0 < d )

3% 0% 7%

3 xo =

0.0084

0% 0% 0% 0%

Totals 8% 3% 0% 11%

Problem 10: A charged particle (q = 4 × 10-10 C) experiences a force of F = 2.5i - 3.8j N in an electric field.


F = 2.5i - 3.8j N

q = 4 × 10-10 C

Part (a) Write an expression for the electric field vector E to which the charge is subject.


E = F/q E = F/q 100%

Grade Detail



Submission Detail


1 E = F/q 0% 0% 0% 0%

Totals 0% 0% 0% 0%

Part (b) Assume this field is generated by a point charge of Q = 5 × 10-9 C. How far away is this charge located? Give your answer in meters.


d = 6.286E-5 d = ( 6.25 * 10-5 )

d = 6.25E-5

100%

Grade Detail



Submission Detail


1 d = ( 6.25 * 10-5 )d = 6.25E-5

0% 0% 0% 0%

Totals 0% 0% 0% 0%

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