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Unit 4 Electric Potential Energy and Electric PotentialPowerPoint Presentation

Unit 4 Electric Potential Energy and Electric Potential

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Unit 4 Electric Potential Energy and Electric Potential

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Unit 4 Electric Potential Energy and Electric Potential

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Unit 4Electric Potential Energy and Electric Potential

Charges and the MVE – WS 4 #1

- In Physics 210 we discussed energy and the MVE.
- This equation still applies to charged particles even though they are so small.
- However, when we discuss charges, we will modify the MVE.
- Gravitational Potential Energy has little impact on charges; therefore, we will ignore it.
- Elastic Potential Energy and Rotational Kinetic Energy will likewise be ignored.
- However, we will need to add a term for Electrical Potential Energy (U).
- This type of energy acts like Gravitational Potential Energy and is caused by the attraction between charged particles.
- The MVE looks like the equation below when these changes are made.

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Electric Potential – WS 4 #2

- In the picture below, the negative charge rest against the positively charged plate, and the positive charge rests against the negatively charged plate.
- Observe what happens as the switches in the picture below are closed.
- Essentially, the electric field was reversed; therefore, the charges moved to the opposite plates.
- In order for the charges to move, there must be a force applied to them that caused them to move through a distance.
- As a result, there was work done on the charges.
- On the next slide we will determine the work done by using the modified MVE.

- In order for the charges in the previous slide to move, work must be done on them.
- Recall that in our mechanical systems, work was determined using the equation below where r is the distance an object is moved.
- From unit 13 we learned that the force on a charged particle in an electric field is given by the following equation.
- As a result, the work done in moving a charge from one position to another in an electric field would be as follows.
- Electrical systems, just as in mechanical systems, can have positive and negative work.
- Consider the two charges below: in which direction do the charges “wish” to move in the electric field shown?
- The negative charge does not wish to move because it is next to the positively charged plate and is “happy.”
- The work done in moving this charge toward the negatively charged plate would be negative because the force on the charge due to the electric field is up whereas the displacement would be down.
- What kind of work was done on the positive charge when we moved it to the negatively charged plate ?
- Explain your answer.

- When we dealt with gravitation potential energy (GPE), the ground was considered to be our lowest point where the height is equal to zero and subsequently the GPE = 0.
- When dealing with charges in uniform electric fields, our reference point where U = 0 would be the charged plate where the electric charge is the “happiest.”
- Is a positive charge “happy” when it is located near a positively charged plate? Why?
- It is not happy because like charges repel each other; therefore, it would have a high potential energy.
- What about a negative charge when it is located near a positively charged plate? Why?
- A negative charge would very “happy” because opposite charges attract; therefore, it would have a zero electric potential energy.
- “Up” or “Down” are meaningless terms in these electrical systems.

- Using the MVE below, derive the equation for the amount of electric potential energy acquired by the negative charge when work is done to move it to the negatively charged plate.
- What are the initial and final kinetic energies of the negatively charged particle?
- What is the initial electric potential energy of the negative charge? Why?
- After eliminating the appropriate terms and making the appropriate substitutions, we find the equation needed to calculate the potential energy of the negative charge when it is located near the negatively charged plate.

- Three charges in an uniform electric field are released from the same distance from a negatively charged metal plate.
- Are they initially in an area of high potential energy or low potential energy? Explain.
- They follow the paths shown.
- Which charge requires the most work in order to move the charge?
- In the end are they in a higher or lower potential energy state? Explain.
- Now consider the red charge.
- How much work is done in moving it from its initial to its final position?
- Explain your answer.

- A charge (q = 8.3 10-8 C) held in place 1.5 cm from a positively charged plate is pushed 0.87 cm closer to the negatively charged plate.
- The magnitude of the electric field is 5.35 105 N/C.
- Calculate the force (magnitude and direction) acting on the charge.
- What is the electric potential energy of the charge in its initial position?
- To find this electric potential energy, use the MVE from the position of zero potential energy to the position it now holds.
- How much work is done in moving the charge as
specified above?

- What is the charges new potential energy?

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- A beam of charged particles (q = -2.2 10-12 C, m = 8.92 10-15) initially traveling at 125.0 m/s is fired through an electric field (E = 900 N/C) generated by the two parallel plates below.
- The particles deflect by 4.5 mm as they pass through the plates (Note: we do not say rise or fall).
- Draw in arrows representing the direction the field is acting in the figure below.
- What is the new speed of the charges as they leave the parallel plates?

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WS 25 # 11 – 13 (Homework)

- Two charges are positioned as below on different plates of two parallel plates below separated by 1.0 cm.
- The purple charge has a charge of |qp| = 2.2 10-12 C and a mass of 5.0 10-11 kg, and the green charge has a charge of |qg| = 1.1 10-12 C and a mass of 5.0 10-11 kg.
- The switches are closed reversing the direction of the electric field, and the charges move as shown.
- What are the charges (+ or -)of the green and purple charges? Explain your answer.
- If the force acting on the purple charge is 1.50 N, then find the following: the magnitude of the electric field, the force acting on the green charge, the work required to move the charges to the opposite plates, and the electric potential energy of the charges immediately before they begin to travel from one plate to the other.
- Calculate the speed of the particles when they collide with the opposite plates.

- On previous slides we derived the equation needed to calculate the potential energy of a charged particle moving in an uniform electric field.
- We will now consider the potential energy of one charged particle in the vicinity of another charged particle.
- Electrical Potential Energy is the energy a charged particle has due to its position relative to an electric field.
- Watch as the test charge, q0, moves from position r1 to position r2 in the electric field generated by the positive charge.
- As this particle moves, its potential energy relative to the large positive charge changes.

Electrical Potential Energy – WS 4 #14 (Homework)

- Electrical Potential Energy is the energy a charged particle has due to its position relative to an electric field.
- Watch as the test charge, q0, moves from position r1 to position r2.
- As this particle moves, its potential energy relative to the large positive charge changes.
- The equation for the particle’s Electrical Potential energy (U) at position 1 and position 2 is as shown to the right.
- The magnitude (+ or -) of the potential energy will depend on the charge of the test charge q0.
- If q0 is positive, then the sign for U will be positive and would increase as q0 moved closer to q1.
- If q0 is negative, then the sign for U will be negative and would decrease as q0 moved closer to q1.
- How would these values change if we replaced q1 in the center with a negative charge?

- Using the MVE below, derive the equation for work required to move the test charge closer to the positive charge.
- What are the initial and final kinetic energies of the test charge?
- Beginning with the MVE and making the appropriate substitutions, we derive the following equation.

WS 4 # 16 & 17

- Two charges (|qbig| = 4.2 10-8 C, and |qsml| = 7.7 10-10 C) are in close proximity to one another in as shown below.
- The red charge is moved from its initial position (r1 = 1.25 cm) to the new position (r2 = 0.17 cm) below.
- Is its new position in an area of higher or lower potential energy? Explain.
- Calculate the work required to move the charge to the new position.
- Now suppose the charge moved along a different path to the same point as shown.
- Without doing any calculations, determine the work needed to move this charge to this same point along the new position.
- Explain your answer.

- Charges q1 and q2 each have a charge of 9.0 10-19 C.
- Each of the squares in the graph are 0.01 cm 0.01 cm.
- The third charge has a charge of q3 = 1.3 10-17 C.
- The charge is moved from point a to point b as shown.
- Has this charge moved to an area of higher or lower potential energy. Explain.
- How much work is done moving this charge from point a to point b?

Electric Potential – WS 4 # 20-21

- Electric Potential (V) is the electric potential energy (U) per unit of charge.
- The equation for electric potential is given below in two forms.
- The units for electric potential is the J/C which is defined as the volt.
- We normally like to consider the Electric Potential Difference (commonly known as voltage) between to different positions.
- On the next slide we will consider the potential difference between two parallel plates.

Electric Potential – WS 4 #22

- Determine the change in electric potential of the charges below when moved to the opposite metal plate.
- What are the initial and final kinetic energies of the charges in the picture below?
- Solving the equation for work, we get the following.
- This work is just the change in the electrical potential energy (U) of the charges.
- Electric Potential (V) is the electric potential energy (U) per unit of charge.
- Substituting, we get the following equation.
- As a result, the electric potential between the starting and ending positions would be as follows.
- This difference in electric potential is known as the voltage between the starting and ending points of the charges as they go from one position to another in the presence of an electric field.

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Electric Potential

- In the previous slide, we derived the equation needed to find the electric potential when we move a charge from an area of zero electric potential energy to and area of higher or from a high electricpotential energy to a zero electric potential energy.
- In this slide we derive the equations needed to determine the electric potential when the charge travels from one non-zero potential energy to another.
- In the animation below both charges move in this manner.

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Electric Potential

- Look at the two charges below.
- The positive charge is very “happy” because it is near the negatively charged plate.
- As a result, we say that the positive charge has U1= 0 (V1= 0) because it does not want to move; however, the negative charge is very “unhappy” because it too is near the negatively charged plate.
- As a result it has a very high U1 > 0 and wants to move away.
- Now lets move the two charges across to the next parallel plate.
- The negative charge has U2= 0 (V2 = 0); however, the positive charge has U2 > 0 (V2 > 0).
- In order to move the charges, we had to do work on them.
- The voltage difference that we had to overcome by the work can be found as follows.

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Why Do “Happy” Charges Move?

- Students often wonder why a charge would move when it is already “happy.”
- Consider the charge below.
- What is the charge on the plates in the figure below?
- The bottom one is negative, and the top one is positive.
- The electric field due to these plates is as shown.
- Watch the charge.
- Why do you think it moved?
- There are many reasons why it might move.
- One possibility is another electric field that
opposes the field established by the AA battery.

- As the flux (electric field density) of this
field is higher, the electric field produced

by the D batteries does more work on the

charge than the field produced by the AA

battery.

- As a result, the charge moves.

- Answer the questions below as they pertain to the charges shown in the figure to the right.
- Which charges have an electric potential V = 0? Explain.
- Do the charges to the right move from a high potential to a low potential or from a low potential to a high potential? Explain.

- What is the change in electric potential of the charge in problem 9?

Electric Potential of Two Point Charges – WS 4 #26

- Derive the equation needed in order to calculate the work done when moving the small charge along the path shown.

- A point charge has a charge of -2.6 10-15 C.
- Find the distances ra, rb, and rc if the electric potentials at these points are 30.0 V, 20.0 V, and 10.0 V respectively.
- What is the electric potential energy of this charge at each of these three points?

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- A small charged body hangs from a string (l = 12.0 cm)in the center between two metal plates that are 3.5 cm apart and connected to a battery as shown below.
- The charge has a charge of 5.7 10-5 C.
- When the switch is closed, the body moves to the position shown where it makes an angle of 15.0 with the vertical.
- What is the magnitude (+ or -) on the charge? Explain.
- How much work is done by the electric field in moving the charge in the horizontal direction.
- What is the magnitude of the electric field?
- What is the change in electric potential of the charge?

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- A small charged body (m = 0.95 g) hangs from a string in the center between two metal plates that are 3.5 cm apart and connected to a battery as shown below.
- The charge has a charge of 5.7 10-5 C.
- When the switch is closed, the body moves to the position shown where it makes an angle of 15.0 with the vertical.
- What is the magnitude (+ or -) on the charge? Explain.
- How much work is done by the electric field in moving the charge in the horizontal direction.
- What is the magnitude of the electric field?
- What is the change in electric potential of the charge?

- We may find the electric potential due to a system of point charges like the one appearing below using the following equation.
- Recall that electric potential is a scalar quantity; therefore, only the distance, not the direction, of the charges from a point in space is needed.
- In the figure below, the square has a length of l = 0.90 m.
- Calculate the electric potential at point P due to the five charges below given that the magnitude of q1 = 15.0 nC and the magnitude of q2 = 22.5 nC.

- An electric charge moves a distance of l from point a to point b through a non-uniform electric field as shown.
- Determine the electric potential difference experienced by this charge.

Equipotential Surfaces

- Equipotential Surfaces indicate positions in an electric field where the electric potential is constant.
- These equipotential surfaces within a uniform electric field are indicated by the blue lines in the figure to the right.
- When a charge moves along the equipotential surfaces, the electric field does no work on the charge.
- The electric field does work on a charge only when it moves from one equipotential surface to another.
- The equipotential surfaces about an isolated point charge would be as shown.

Equipotential Surfaces for a Dipole WS 4 #37

- The equipotential surfaces for a dipole would look those appearing in the figure to the right.
- Remember, if a charge moves along these surfaces, then the electric field does no work on the charge.
- Only when a charge moves from one surface to the other does the electric field do work on a charge.

- Consider the two points P1 and P2 separated by a distance of l and located a distance of r1 and r2 from point P.
- If r1>>l and r2>>l, then we can approximate that r1 r2.
- As a result, the difference in the lengths of r1 and r2 may be found as follows.
- Additionally, if l is very small, then the following is also true.

- Calculate the electric potential at point P due to the dipole below given that the magnitude of q1 = q2 = 15.0 nC = q.

- When we have a long rod with a continuous charge distribution, we need to integrate over the charge distribution in order to find the electric potential at point P.

- Determine the electric potential at point P located a distance of z from a uniformly charged disk of radius R.

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- Determine the electric potential at point P located a distance of z from a uniformly charged disk of radius R.