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Electricity & Lecture 5:

Today... ! Electric Potenal Energy

! “I noced the formula sheet does not have the potenal energy equaon used in the prelecture. Will this be changed on the sheet we will use for exams or are we supposed to be able to ﬁnd this equaon through the given equaon?”

! You should be able to ﬁgure out Potenal Energy from the Electric Potenal and the principle of superposion.

! “How much harder does this course get down the line? This course already has me shook.”

! It doesn’t necessarily get easy but most people make it through. Gauss’ Law is usually puzzling at ﬁrst.

Electricity & Magnesm Lecture 5, Slide 2 CheckPoint Result: EPE of Point Charge

A charge of +Q is ﬁxed in space. A charge of +q was ﬁrst placed at a distance r1 away from +Q. Then it was moved along a straight line to a new posion at a distance R away from its starng posion. The ﬁnal locaon of +q is at a distance r2 from +Q. What is the change in the potenal energy of charge +q during this process? A. kQq/R 2 B. kQqR/r1 2 C. kQqR/r2

D. kQq((1/r2)-(1/r1))

E. kQq((1/r1)-(1/r2))

“kQq/r is the potenal at a point so the diﬀerence in these two potenals will be found by doing kQq((1/r2)-(1/r1))”

“Since the parcle is moved away from the ﬁxed charge, the potenal energy must increase. The part 1/r1-1/r2 would yield a posive answer because 1/r1>1/r2”

Electricity & Magnesm Lecture 5, Slide 12 CheckPoint Result: EPE of Point Charge

A charge of +Q is ﬁxed in space. A second charge of +q was ﬁrst placed at a distance r1 away from +Q. Then it was moved along a straight line to a new posion at a distance R away from its starng posion. The ﬁnal locaon of +q is at a distance r2 from +Q. What is the change in the potenal energy of charge +q during this process?

A. kQq/R 2 B. kQqR/r1 2 C. kQqR/r2

D. kQq((1/r2)-(1/r1))

E. kQq((1/r1)-(1/r2))

Qq 1 1 U U U = ⌘ 2 1 4⇡" r r 0 2 1 ! Note: +q moves AWAY from +Q. Its Potenal energy MUST DECREASE ΔU < 0

Electricity & Magnesm Lecture 5, Slide 13 Recall from Mechanics:

~r2 W = F~ d~r · ~r1 R F dr W > 0 Object speeds up ( ΔK > 0 )

F dr or F W < 0 Object slows down ( ΔK < 0 ) dr

F dr W = 0 Constant speed ( ΔK = 0 )

Electricity & Magnesm Lecture 5, Slide 3 Dot Product (review)

If you know A and B in rectangular coordinates, then you can ﬁnd the angle between them.

If does negave , potenal energy increases! Same idea for force… if Coulomb force does negave work, potenal energy increases.

+ + F

Δx Coulomb force does negave work + + Potenal energy increases

Electricity & Magnesm Lecture 5, Slide 4 CheckPoint: of Point Charge Electric

A charge is released from rest in a region of electric ﬁeld. The charge will start to move

A) In a direcon that makes its potenal energy increase. B) In a direcon that makes its potenal energy decrease. C) Along a path of constant potenal energy.

It will move in the same direcon as F

F Work done by force is posive

Δx ΔU = −Work is negave

Nature wants things to move in such a way that PE decreases

Electricity & Magnesm Lecture 5, Slide 5 Clicker Question

You hold a posively charged ball and walk due west in a region that contains an electric ﬁeld directed due east.

FE FH East dr West

WH is the work done by the hand on the ball

WE is the work done by the electric ﬁeld on the ball

Which of the following statements is true:

A) WH > 0 and WE > 0

B) WH > 0 and WE < 0

C) WH < 0 and WE < 0

D) WH < 0 and WE > 0 Electricity & Magnesm Lecture 5, Slide 6 Clicker Question

Not a conservave force. Conservave force: ΔU = − WE Does not have any ΔU.

FE FH E dr

B) WH > 0 and WE < 0

Is ΔU posive or negave?

A) Posive

B) Negave

Electricity & Magnesm Lecture 5, Slide 7 Checkpoint

Masses M1 and M2 are inially separated by a distance Ra. M2 is now moved further away from mass M1 such that their ﬁnal separaon is Rb. Which of the following statements best describes the work Wab done by the force of gravity on M2 as it moves from Ra to Rb?

A) Wab > 0 B) Wab = 0 C) Wab < 0 Example: Two Point Charges

Calculate the change in potenal energy for two point charges originally very far apart moved to a separaon of “d”

d q 1 q2

� is another epsilon Charged parcles with the same sign have an increase in potenal energy when brought closer together.

For point charges oen choose r = inﬁnity as “zero” potenal energy.

Electricity & Magnesm Lecture 5, Slide 8 Clicker Question

Case A d

Case B 2d

In case A two negave charges which are equal in magnitude are separated by a distance d. In case B the same charges are separated by a distance 2d. Which conﬁguraon has the highest potenal energy?

A) Case A B) Case B

Electricity & Magnesm Lecture 5, Slide 9 Clicker Question Discussion

As usual, choose U = 0 to be at inﬁnity:

Case A d Case B 2d U(r)

UA > UB U(d) U(2d)

0 r

Electricity & Magnesm Lecture 5, Slide 10 Checkpoint

In Case A two charges which are equal in magnitude but opposite in charge are separated by a distance d. In Case B the same charges are separated by a distance 2d. Which conﬁguraon has the highest potenal energy U?

A) Case A has the highest potenal energy

B) Case B has the highest potenal energy

C) Both cases have the same potenal energy

Potential Energy of Many Charges

Two charges are separated by a distance d. What is the change in potenal energy when a third charge q is brought from far away to a distance d from the original two charges?

Q2

d d

(superposion) Q1 q d

Electricity & Magnesm Lecture 5, Slide 14 Potential Energy of Many Charges What is the total energy required to bring in three idencal charges, from inﬁnitely far away to the points on an equilateral triangle shown.

A) 0 Q B)

C) d d D)

E) Q d Q

Work to bring in ﬁrst charge: W1 = 0 Work to bring in second charge :

Work to bring in third charge :

Electricity & Magnesm Lecture 5, Slide 15 Potential Energy of Many Charges

Suppose one of the charges is negave. Now what is the total energy required to bring the three charges in inﬁnitely far away? Q 2 A) 0 B) d d C) D) Q Q E) 1 d

Work to bring in ﬁrst charge: W1 = 0 Work to bring in second charge :

Work to bring in third charge :

Electricity & Magnesm Lecture 5, Slide 16 CheckPoint: EPE of a of Point Charges 1

Two charges which are equal in magnitude, but opposite in sign, are placed at equal distances from point A as shown. If a third charge is added to the system and placed at point A, how does the electric potenal energy of the charge collecon change? A. Potenal energy increases B. Potenal energy decreases C. Potenal energy does not change D. The answer depends on the sign of the third charge

If the new charge has a potenal energy caused by charge 1 equal to charge 2, when we sum them up, it will add up to 0, resulng in no change in total potenal energy. third charge will add kq1q3/d and kq2q3/d to the total U, unless q3 is zero, it will deﬁnitely change the total potenal energy of the system.

Electricity & Magnesm Lecture 5, Slide 17 Question

“Could you go over the third checkpoint, in some detail and show how the calculaon would look like for this kind of problem” CheckPoint: EPE of a System of Point Charges 2

Two point charges are separated by some distance as shown. The charge of the ﬁrst is posive. The charge of the second is negave and its magnitude is twice as large as the ﬁrst. Is it possible ﬁnd a place to bring a third charge in from inﬁnity without changing the total potenal energy of the system?

A. YES, as long as the third charge is posive B. YES, as long as the third charge is negave C. YES, no maer what the sign of the third charge D. NO HOW? LET’S DO THE CALCULATION! It is possible to bring a third charge in from inﬁnity without changing the total potenal energy of the system if this third charge is placed a distance twice as far way from the negave charge than from the posive charge (and q1 and q2 must be equal in magnitude?).

“The potential the third charge will contribute to the system have opposite signs but NOT equal magnitudes. So the net potential energy will not equal 0.”

Electricity & Magnesm Lecture 5, Slide 18 Example

A posive charge q is placed at x = 0 and a negave charge −2q is placed at x = d. At how many diﬀerent places along the x axis could another posive charge be placed without changing the total potenal energy of the system?

q −2q x X = 0 X = d A) 0 B) 1 C) 2 D) 3

Electricity & Magnesm Lecture 5, Slide 19 Example

At which two places can a posive charge be placed without changing the total potenal energy of the system?

q −2q A x X = 0 B C X = d D

A) A & B B) A & C C) B & C Let’s calculate the posions of A and B D) B & D E) A & D

Electricity & Magnesm Lecture 5, Slide 20 Lets work out where A is

r d

q −2q A x X = 0 X = d

Set ΔU = 0

Makes Sense! Q is twice as far from −2q as it is from +q

Electricity & Magnesm Lecture 5, Slide 21 Lets work out where B is

r d − r

q −2q x X = 0 B X = d

Seng ΔU = 0

Makes Sense! Q is twice as far from −2q as it is from +q

Electricity & Magnesm Lecture 5, Slide 22 What about D?

Can you prove that is not possible to put another charge at posion D without changing U?

d r

q −2q x X = 0 X = d D

1 2 (r must be posive) r+d = r Summary

For a pair of charges: r

Just evaluate q1 q2

(We usually choose U = 0 to be where the charges are far apart)

For a collecon of charges:

Sum up for all pairs

Electricity & Magnesm Lecture 5, Slide 23