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NAME________Solutions__________
Physics 114 โ Exam 1 Section A (MWF)
September 20th, 2017
There are 3 parts to the exam including multiple choice, short answer, and calculation-based problems. Please note that in parts II and III, you can skip one of the questions.
Score: Part I[ ____/20] + Part II[ ____/20] + Part III[ ____/40] = Total[ ______/80]
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. What determines the direction of the electric field just outside the surface of a conductor?
a. It will be parallel to the surface and proportional to the potential b. It will be parallel to the surface and proportional to the surface charge c. It will be perpendicular to the surface and proportional to the potential d. It will be perpendicular to the surface and proportional to the surface charge e. The electric field always vanishes just outside a conductor
2. In the figure below, what is the magnitude of the electrostatic force on ๐, if
๐ = 30 ๐๐ถ, ๐ = 5 ๐๐ถ, and ๐ = 30๐๐. ๐น!"! =!!!"!!
โ !!!!"!!!
= !!!"!!
!!= 7.5๐
a. ๐.๐ ๐ต b. 38 ๐ c. 15 ๐ d. 23 ๐ e. ๐ง๐๐๐ 3. Which of the following units is appropriate for measuring electric potential? a. N/C (only) b. V/m (only) c. N/C or V/m d. F (only) e. V (only) 4. A system consisting of a negatively-charged particle and an electric field a. Gains potential difference and kinetic energy when the charged particle moves in the direction of the field b. Loses electric potential energy when the charged particle moves in the direction of the field c. Gains kinetic energy when the charged particle moves in the direction of the field d. Gains electric potential energy when the charged particle moves in the direction of the field e. None of the above
5. A cube of edge a has a charge of 12 ๐๐ถ in its exact center, and 6 additional
charges, each of magnitude โ2 ๐๐ถ spaced a distance a away from each face of the cube (such that each additional charge is outside the box.) What is the total electric flux out of the cube?
a. 0 ๐๐ถ/๐! b. 24 ๐๐ถ/๐! c. 8 ๐๐ถ/๐! d. ๐๐ ๐๐ช/๐๐ e. โ24๐๐ถ/๐!
6. A positive point charge q is placed at the center of an
uncharged metal sphere insulated from the ground. The outside of the sphere is then grounded (as shown). Then the ground wire is removed. A is the inner surface and B is the outer surface. Which statement is correct?
a. The charge on A is โq; that on B is +q b. The charge on B is โq; that on A is +q c. The charge is q/2 on A and on B d. There is no charge on either A or B e. The charge on A is โq; there is no charge on B.
7. A carbon monoxide molecule has a negatively charged
oxygen atom connected to a positively charged carbon atom, roughly sketched in the figure. The total charge is zero. If an electric field is present pointing to the left, a carbon monoxide molecule will feel what sort of force on it?
a. A net force opposite the direction of the field b. A net force in the direction of the field c. No net force or torque d. A torque, or twisting force, that tries to get the oxygen atom to the right e. A torque, or twisting force, that tries to get the carbon atom to the right
8. Suppose a conducting sphere has a large, positive charge placed on it by
connecting it to a battery. Which of the following has occurred? a. Protons have been added b. Electrons have been added c. Protons have been removed d. Electrons have been removed e. None of the above
9. A student has made the claim that the electric flux through one half of a Gaussian
surface is always equal to the flux through the other half of the surface. This is a. Never true b. Always true c. True whenever the enclosed charge is symmetrically located at a center point, on
a center line, or a centrally placed plane d. True whenever no charge is enclosed within the Gaussian surface e. True only when no charge is enclosed within the Gaussian surface
10. Where on the surface of a conductor will the electric field generally be the
strongest? a. The point on the conductor closest to any external positive charge b. The point on the conductor closest to the center c. The point of the conductor furthest from the center d. Where the conductor has low curvatures, such as where it is flat e. Where the conductor has high curvature, such as a sharp point
Part II: Short Answer [20 points]
Choose two of the following questions and give a short answer (2-4 sentences) and/or brief sketch (10 points each).
1. Sketch a basic picture and describe how the Millikan Oil Drop experiment works.
โข Atomizer produced tiny drops of oil; gravity pulls them down
โข Atomizer induces small charges โข E-field opposes gravity โข Can tune the E-field to exactly oppose
gravity โข He found that all charges on the oil drops
were integer multiples of the fundamental charges
2. The points in the figure below are on a series of equipotential surfaces associated with an electric field. Rank (from greatest to least) the work done by the electric field on a positively charged particle moving from A to B, from B to C, from C to D, and from D to E. Explain your reasoning.
๐ต โ ๐ถ > ๐ถ โ ๐ท > ๐ด โ ๐ต > ๐ท โ ๐ธ
Here, we have A and B on the same 9V surface, C and E are on the 7 V surface, and D is on the lowest, 6V surface. The work done by the electric field is
๐ = โฮ๐, ฮ๐ โก !!!โ ๐ = โ๐ฮ๐. Therefore,
๐ต โ ๐ถ: ๐! โ ๐! = ฮ๐ = 7๐ โ 9๐ = โ2๐, ๐ = โ๐ฮ๐ = ๐ 2๐
๐ถ โ ๐ท: ๐! โ ๐! = ฮ๐ = 6๐ โ 7๐ = โ1๐, ๐ = โ๐ฮ๐ = ๐ 1๐
๐ด โ ๐ต: ๐! โ ๐! = ฮ๐ = 9๐ โ 9๐ = 0๐, ๐ = โ๐ฮ๐ = 0
๐ท โ ๐ธ: ๐! โ ๐! = ฮ๐ = 7๐ โ 6๐ = 1๐, ๐ = โ๐ฮ๐ = โ๐ 1๐
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3. On the figure below, sketch in the approximate equipotential lines. Include 3 or more curves, and indicate which ones represent the highest potential and which represent the lowest potential.
The equipotential lines are crudely sketched in below. Equipotential surfaces are always be perpendicular to the electric field lines. The highest ones are around the charge A and the lowest ones around charge B. Donโt take my sketches too literally.
Part III: Calculation [40 points]
Choose two of following three questions and perform the indicated calculations. You must show your work for full credit. (20 points each)
1. Four point charges are arranged in the xy-plane as sketched below. Two charges have magnitude +8 ๐๐ถ and two have โ8 ๐๐ถ. They are all located a distance of ๐ = 4๐๐ from the origin.
a. Find the magnitude of the electric field at the origin. b. Now, a point charge of ๐ = +1.00๐๐ถ and mass ๐ = 1.00๐ is placed at the origin.
Find the force on the charge (magnitude and direction) and the acceleration of the point charge (magnitude and direction).
This is a vector summing problem. We want to find the x and y components of the E-field from each point charge at the place of interest (the origin) and then combine the results
๐ธ =๐!๐๐!
๐
Letโs label the points A, B, C, D starting with the top charge as A and going clockwise to end with the leftmost charge as D.
๐ธ! =๐!๐๐!
+๐ฆ , ๐ธ! =๐!๐๐!
โ๐ฅ , ๐ธ! =๐!๐๐!
+๐ฆ , ๐ธ! =๐!๐๐!
โ๐ฅ
๐ธ! =2๐!๐๐!
โ๐ฅ , ๐ธ! =2๐!๐๐!
+๐ฆ
๐ธ! = ๐ธ! =2 8.988ร10! ๐๐!
๐ถ! 8ร10!! ๐ถ
0.04 ๐! = 8.988ร10!
๐๐ถ
๐ธ = ๐ธ!! + ๐ธ!! = 1.27ร10!๐๐ถ
๐ = tan!!๐ธ!๐ธ!
= โ45โ โ 180โ โ 45โ = 135โ ๐ถ๐ถ๐ ๐๐๐๐ ๐๐๐๐๐๐
Now the force will be in the same direction of ๐ธ for positive charge and is ๐น = ๐๐ธ,
๐น = +1๐๐ถ 1.27ร10!๐๐ถ
= 127 ๐
And ๐น = ๐๐, so, again the direction is the same and we have
๐ =๐น๐=
127๐1ร10!!๐๐
= 1.27ร10!๐๐ !
โ8๐๐ถ
+8๐๐ถ
โ8๐๐ถ +8๐๐ถ
๐
2. The potential in a certain region of space is given by ๐ = โ๐ต๐ฆ๐ง, where ๐ต = 5ร10! !!!.
a. Find all three components of the electric field, ๐ธ! ,๐ธ! , and ๐ธ! as functions of ๐ฆ and
๐ง. b. A particle of mass ๐ = 4.00๐ and charge ๐ = 2 ๐๐ถ is initially at rest at position
๐ฅ, ๐ฆ, ๐ง = 0.00, 1.00, 1.00 ๐๐. Find the initial acceleration of this charge. c. The particle shortly thereafter finds itself at ๐ฅ, ๐ฆ, ๐ง = 0.00, 2.00, 2.00 ๐๐. How
much did its potential energy change?
๐ธ = โโ๐ = โ๐๐!๐๐ฅ
๐ฅ +๐๐!๐๐ฆ
๐ฆ +๐๐!๐๐ง
๐ง
๐ธ! = โ๐๐!๐๐ฅ
= 0
๐ธ! = โ๐๐!๐๐ฆ
= ๐ต๐ง
๐ธ! = โ๐๐!๐๐ง
= ๐ต๐ฆ
Now a particle is at (0,1,1) and is at rest. We can find the acceleration from the
force, ๐น = ๐๐ธ = ๐๐. We will also convert the units, since !!= !
!
๐น! = ๐๐ธ! = 0
๐น! = ๐๐ธ! = ๐๐ต๐ง = 2ร10!!๐ถ 5ร10!๐๐!
๐๐ถ โ ๐
0.01๐ = 0.1๐
๐น! = ๐๐ธ! = ๐๐ต๐ฆ = 2ร10!!๐ถ 5ร10!๐๐!
๐๐ถ โ ๐
0.01๐ = 0.1๐
๐ =๐น๐=0.1๐ฅ + 0.1๐0.004๐๐
๐ = 25 ๐ฅ + ๐๐๐ !
Now it moves to a new position ๐ฅ, ๐ฆ, ๐ง = 0,2,2 ๐๐ and we want to find the change in PE and using its relationship to the potential ๐ = โ๐ต๐ฆ๐ง,
ฮ๐ = ๐! โ ๐! = ๐ฮ๐ = ๐ ๐! โ ๐! ฮ๐ = โ๐ ๐ต 0.02 0.02 โ ๐ต 0.01 0.01 = โ3ร10!!๐ต๐ ๐!
ฮ๐ = โ3ร10!! 5ร10! ๐
๐ถ โ ๐ 2ร10!!๐ถ ๐!
ฮ๐ = โ0.003 ๐ฝ
3. Find the net electric flux for the following cases: a. A square surface of area ๐ด = 2๐! that is parallel to the
๐ฅ๐ง โplane that is in a constant electric field given by ๐ธ = 5๐ค +5๐ฅ !
!
b. The closed spherical surface with ๐ = 2๐ in a uniform electric
field of magnitude ๐ธ = 10 !!
as shown in the figure
c. The closed cylindrical surface as shown below where the
magnitude of the electric field is ๐ธ = 10 !!
and points out
of the surface on both sides as shown below. For the cylinder, ๐ = 2๐ and the length is ๐ฟ = 5๐.
d. What can you conclude about the charges, if any, inside the cylindrical surface (of part c)?
The first portion is a square surface and we can solve this using Cartesian coordinates or in terms of the magnitudes of the vectors and the angles between them. The angle is
๐ = 45โ. The magnitude of the electric field is ๐ธ = ๐ธ!! + ๐ธ!! = 50 = 7.07 !!
ฮฆ = ๐ธ โ ๐ด = ๐ธ ๐ด cos ๐ = 7.07๐๐ถ
2 ๐! cos๐4
= 10๐๐ถ๐!
ฮฆ = ๐ธ โ ๐ด = 5๐ค + 5๐ฅ 2๐ฅ = 10๐๐ถ๐!
For part B, we have a sphere in a uniform field. Because all the field lines that go into the sphere come out the other side, there is no net flux.โด ฮฆ!"# = 0
In the last portion, no flux goes through the lateral surface, and the same amount of E-field lines go out of the right side as the left. So we can find it for one surface and multiply by 2. Also, we know that ๐ธ is in the same direction as ๐, and the area of a circle is ๐ด = ๐๐ ! so
ฮฆ = 2๐ธ๐ด cos ๐ = 2๐ธ๐ด = 2 10๐๐ถ
๐ 2!(๐! )
ฮฆ = 80๐๐๐ถ๐! = 251
๐๐!
๐ถ
Now, what can we say about the charges inside? Well we have a flux out, so we have charge inside given by,
๐!"# = ฮฆ๐! = ฮฆ1
4๐๐!= 251
๐๐ถ๐! 1
4๐๐!๐ถ!
๐ โ ๐!
๐!"# = 2.22๐๐ถ
Potentially Useful Information
Constants (units)
๐ = 1.602ร10!!"(๐ถ)
๐! = 8.988ร10! !๐ โ ๐!
๐ถ!!
๐!! = 9.109ร10!!"(๐๐) ๐! = 1.673ร10!!"(๐๐)
Cylinders
๐ = ๐๐ !๐ฟ
๐ด!"# = 2๐๐ ๐ฟ
Spheres
๐ =43๐๐ !
๐ด = 4๐๐ !
Circles
๐ด = ๐๐ !
๐ถ = 2๐๐
Metric Prefixes
๐ = 10!!, ๐ = 10!!
๐ = 10!!, ๐ = 10!!"
Triangles
๐ด =12๐ต๐ป
๏ฟฝโ๏ฟฝ = ๐๏ฟฝโ๏ฟฝ
๐พ๐ธ =12๐๐ฃ!
๐ธ =๐2๐!
!ยฑ๐!!