NAME________Solutions__________

Physics 114 – 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 𝑞, if

𝑄 = 30 𝜇𝐶, 𝑞 = 5 𝜇𝐶, and 𝑑 = 30𝑐𝑚. 𝐹!"! =!!!"!!

− !!!!"!!!

= !!!"!!

!!= 7.5𝑁

a. 𝟕.𝟓 𝑵 b. 38 𝑁 c. 15 𝑁 d. 23 𝑁 e. 𝑧𝑒𝑟𝑜 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 𝑛𝐶 in its exact center, and 6 additional

charges, each of magnitude −2 𝑛𝐶 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 𝑛𝐶/𝜖! b. 24 𝑛𝐶/𝜖! c. 8 𝑛𝐶/𝜖! d. 𝟏𝟐 𝒏𝑪/𝝐𝒐 e. −24𝑛𝐶/𝜖!

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 –q; that on B is +q b. The charge on B is –q; 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 –q; 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.

• Atomizer produced tiny drops of oil; gravity pulls them down

• Atomizer induces small charges • E-field opposes gravity • Can tune the E-field to exactly oppose

gravity • 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.

𝐵 → 𝐶 > 𝐶 → 𝐷 > 𝐴 → 𝐵 > 𝐷 → 𝐸

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

𝑊 = −Δ𝑈, Δ𝑉 ≡ !!!→ 𝑊 = −𝑞Δ𝑉. Therefore,

𝐵 → 𝐶: 𝑉! − 𝑉! = Δ𝑉 = 7𝑉 − 9𝑉 = −2𝑉, 𝑊 = −𝑒Δ𝑉 = 𝑒 2𝑉

𝐶 → 𝐷: 𝑉! − 𝑉! = Δ𝑉 = 6𝑉 − 7𝑉 = −1𝑉, 𝑊 = −𝑒Δ𝑉 = 𝑒 1𝑉

𝐴 → 𝐵: 𝑉! − 𝑉! = Δ𝑉 = 9𝑉 − 9𝑉 = 0𝑉, 𝑊 = −𝑒Δ𝑉 = 0

𝐷 → 𝐸: 𝑉! − 𝑉! = Δ𝑉 = 7𝑉 − 6𝑉 = 1𝑉, 𝑊 = −𝑒Δ𝑉 = −𝑒 1𝑉

-

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. Don’t 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 𝜇𝐶 and two have −8 𝜇𝐶. They are all located a distance of 𝑎 = 4𝑐𝑚 from the origin.

a. Find the magnitude of the electric field at the origin. b. Now, a point charge of 𝑞 = +1.00𝜇𝐶 and mass 𝑚 = 1.00𝑔 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

𝐸 =𝑘!𝑞𝑟!

𝑟

Let’s 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.

𝐸! =𝑘!𝑞𝑎!

+𝑦 , 𝐸! =𝑘!𝑞𝑎!

−𝑥 , 𝐸! =𝑘!𝑞𝑎!

+𝑦 , 𝐸! =𝑘!𝑞𝑎!

−𝑥

𝐸! =2𝑘!𝑞𝑎!

−𝑥 , 𝐸! =2𝑘!𝑞𝑎!

+𝑦

𝐸! = 𝐸! =2 8.988×10! 𝑁𝑚!

𝐶! 8×10!! 𝐶

0.04 𝑚! = 8.988×10!

𝑁𝐶

𝐸 = 𝐸!! + 𝐸!! = 1.27×10!𝑁𝐶

𝜃 = tan!!𝐸!𝐸!

= −45∘ → 180∘ − 45∘ = 135∘ 𝐶𝐶𝑊 𝑓𝑟𝑜𝑚 𝑜𝑟𝑖𝑔𝑖𝑛

Now the force will be in the same direction of 𝐸 for positive charge and is 𝐹 = 𝑞𝐸,

𝐹 = +1𝜇𝐶 1.27×10!𝑁𝐶

= 127 𝑁

And 𝐹 = 𝑚𝑎, so, again the direction is the same and we have

𝑎 =𝐹𝑚=

127𝑁1×10!!𝑘𝑔

= 1.27×10!𝑚𝑠!

−8𝜇𝐶

+8𝜇𝐶

−8𝜇𝐶 +8𝜇𝐶

𝑎

2. The potential in a certain region of space is given by 𝑉 = −𝐵𝑦𝑧, where 𝐵 = 5×10! !!!.

a. Find all three components of the electric field, 𝐸! ,𝐸! , and 𝐸! as functions of 𝑦 and

𝑧. b. A particle of mass 𝑚 = 4.00𝑔 and charge 𝑞 = 2 𝜇𝐶 is initially at rest at position

𝑥, 𝑦, 𝑧 = 0.00, 1.00, 1.00 𝑐𝑚. Find the initial acceleration of this charge. c. The particle shortly thereafter finds itself at 𝑥, 𝑦, 𝑧 = 0.00, 2.00, 2.00 𝑐𝑚. How

much did its potential energy change?

𝐸 = −∇𝑉 = −𝜕𝑉!𝜕𝑥

𝑥 +𝜕𝑉!𝜕𝑦

𝑦 +𝜕𝑉!𝜕𝑧

𝑧

𝐸! = −𝜕𝑉!𝜕𝑥

= 0

𝐸! = −𝜕𝑉!𝜕𝑦

= 𝐵𝑧

𝐸! = −𝜕𝑉!𝜕𝑧

= 𝐵𝑦

Now a particle is at (0,1,1) and is at rest. We can find the acceleration from the

force, 𝐹 = 𝑞𝐸 = 𝑚𝑎. We will also convert the units, since !!= !

!

𝐹! = 𝑞𝐸! = 0

𝐹! = 𝑞𝐸! = 𝑞𝐵𝑧 = 2×10!!𝐶 5×10!𝑉𝑚!

𝑁𝐶 ⋅𝑚

0.01𝑚 = 0.1𝑁

𝐹! = 𝑞𝐸! = 𝑞𝐵𝑦 = 2×10!!𝐶 5×10!𝑉𝑚!

𝑁𝐶 ⋅𝑚

0.01𝑚 = 0.1𝑁

𝑎 =𝐹𝑚=0.1𝚥 + 0.1𝑘0.004𝑘𝑔

𝑎 = 25 𝚥 + 𝑘𝑚𝑠!

Now it moves to a new position 𝑥, 𝑦, 𝑧 = 0,2,2 𝑐𝑚 and we want to find the change in PE and using its relationship to the potential 𝑉 = −𝐵𝑦𝑧,

Δ𝑈 = 𝑈! − 𝑈! = 𝑞Δ𝑉 = 𝑞 𝑉! − 𝑉! Δ𝑈 = −𝑞 𝐵 0.02 0.02 − 𝐵 0.01 0.01 = −3×10!!𝐵𝑞 𝑚!

Δ𝑈 = −3×10!! 5×10! 𝑁

𝐶 ⋅𝑚 2×10!!𝐶 𝑚!

Δ𝑈 = −0.003 𝐽

3. Find the net electric flux for the following cases: a. A square surface of area 𝐴 = 2𝑚! that is parallel to the

𝑥𝑧 −plane that is in a constant electric field given by 𝐸 = 5𝚤 +5𝚥 !

!

b. The closed spherical surface with 𝑅 = 2𝑚 in a uniform electric

field of magnitude 𝐸 = 10 !!

as shown in the figure

c. The closed cylindrical surface as shown below where the

magnitude of the electric field is 𝐸 = 10 !!

and points out

of the surface on both sides as shown below. For the cylinder, 𝑅 = 2𝑚 and the length is 𝐿 = 5𝑚.

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

𝜃 = 45∘. The magnitude of the electric field is 𝐸 = 𝐸!! + 𝐸!! = 50 = 7.07 !!

Φ = 𝐸 ⋅ 𝐴 = 𝐸 𝐴 cos 𝜃 = 7.07𝑁𝐶

2 𝑚! cos𝜋4

= 10𝑁𝐶𝑚!

Φ = 𝐸 ⋅ 𝐴 = 5𝚤 + 5𝚥 2𝚥 = 10𝑁𝐶𝑚!

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.∴ Φ!"# = 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 𝐸 is in the same direction as 𝑛, and the area of a circle is 𝐴 = 𝜋𝑅! so

Φ = 2𝐸𝐴 cos 𝜃 = 2𝐸𝐴 = 2 10𝑁𝐶

𝜋 2!(𝑚! )

Φ = 80𝜋𝑁𝐶𝑚! = 251

𝑁𝑚!

𝐶

Now, what can we say about the charges inside? Well we have a flux out, so we have charge inside given by,

𝑞!"# = Φ𝜖! = Φ1

4𝜋𝑘!= 251

𝑁𝐶𝑚! 1

4𝜋𝑘!𝐶!

𝑁 ⋅𝑚!

𝑞!"# = 2.22𝑛𝐶

Potentially Useful Information

Constants (units)

𝑒 = 1.602×10!!"(𝐶)

𝑘! = 8.988×10! !𝑁 ⋅𝑚!

𝐶!!

𝑚!! = 9.109×10!!"(𝑘𝑔) 𝑚! = 1.673×10!!"(𝑘𝑔)

Cylinders

𝑉 = 𝜋𝑅!𝐿

𝐴!"# = 2𝜋𝑅𝐿

Spheres

𝑉 =43𝜋𝑅!

𝐴 = 4𝜋𝑅!

Circles

𝐴 = 𝜋𝑅!

𝐶 = 2𝜋𝑅

Metric Prefixes

𝑚 = 10!!, 𝜇 = 10!!

𝑛 = 10!!, 𝑝 = 10!!"

Triangles

𝐴 =12𝐵𝐻

�⃗� = 𝑚�⃗�

𝐾𝐸 =12𝑚𝑣!

𝐸 =𝜎2𝜖!

!±𝑘!!