Ch24 potential

Rogers: Lectures based on Halliday, Resnick and Walker’s Fundamentals of Physics, Copyright 2005 by Wiley and Sons

Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to Copyright Protection. 1

Ch24-1/57

If a force is conservative

=> we can associate a potential energy with it.

=> principle of conservation of mechanical energyapplies in closed systems involving the force.

Review ch 7 and 8.

-change in potentialenergy: W is the work done on the particles by the applied force.

-change in kineticenergy

-work by a constant force

Ch 24: Electric Potential HRW628-645

same direction =>+ve work

Ch24-2/57

Assign electric potential energy U to a system of two or more particles with electrostatic forces between them.

From Ch 8, as system changes from an initial state, i, to a final state, f, the electrostatic force does work on the particles and

The work done by a conservative force is path independent (assumes rest of the system doesn’t change)

Electric Potential Energy

Potential energy requires a reference configurationand a reference potential energy be set/defined.

Usually take reference condition as all particles infinitely separated and define this to have 0 potential energy.

(Force does +ve work => potential decreases)



Ch24-3/57

An electric potential energy is associated with a system of particles as a whole.

But we often speak of the potential energy of a single particle - this is common usage but recall we are really speaking of the potential energy of the entire system.

We sometimes speak of the work done on a single particle by the electric field, i.e. by the force due to the charges that set up the field.

Electric potential energy: Work done by a field

Ch24-4/57

Key ideas: i) change in potential is related to work done on electron

ii) force is constant so

iii)

and so

i.e. the potential energy of the e- decreases.

Electric potential energy: work done by a fieldEarth’s electric field (in the atmosphere) is downwards near the surface. Magnitude is E = 150 N/C. When a cosmic ray knocks an electron out of an atom, what is the change in its electric potential, ∆U, when it goes upwards by 520 m?



Ch24-5/57

A proton moves from i to f in a uniform electric field E.

a) Does the field do +ve or -ve work on the proton?

b) Does the proton’s electric potential increase or decrease?

Question

Ch24-6/57

A proton moves from i to f in a uniform electric field E.

a) Does the field do +ve or -ve work on the proton?

b) Does the proton’s electric potential increase or decrease?

Question

Ans: a)

W = qEd cos180o = -qEd and q > 0 => W is -ve

(i.e. external force has to work to do it).

Ans b) so the potential increases

> 0



Ch24-7/57

If work is positive, this means force helps movement in direction of d.

If work is negative, the force has opposed movement in direction of d.

Gravity helps a ball roll down a hill. Wgravity > 0

Gravity opposes a ball roll up a hill. Wgravity < 0

The sign of work

Ch24-8/57

The electric potential V for any point in an electric field is defined as the potential energy per unit charge.

The electric potential, or just potential, at a point is a scalar, not a vector.

The electric potential difference between any two points in the field, i and f is

or, since ∆U=-W

Electric potential



Ch24-9/57

Potential difference between two points is the negativeof the work done by the electrostatic force to move a unit charge from one point to another. Its sign depends on signs and magnitudes of q and W.

Let Ui = 0 at infinity as our reference potential energy. V must be 0 there too since V=U/q.

We define electric potential at any point in the electric field as

where is work done by the electric field on a charged particle as that particle moves from infinity to a point f.

Electric potential difference

Ch24-10/57

The SI unit is J/C which is given the special name volt, V (cap since named after a person).

1 volt = 1 joule per coulomb

Potential is +ve => work done by the electrostatic force to bring a +ve charge to that point is -ve

=> the electrostatic forces opposed bringing it there.

For a -ve charge the work would be +ve => the electrostatic forces helped bring the charge there.

Potential is +ve => +ve charge wants to leave

=> -ve charge is attracted

Electric potential difference

Count Alessandro Volta Italian 1745-1827



Ch24-11/57

We have been using N/C as the unit for an electric field.

There is 1V/(1J/C) (from defn we just gave). Definition of a joule implies there is 1 J/(1N m) (work = force x distance). Hence

So a natural unit for electric fields is V/m (= N/C)

The electron-volt (eV)

One eV is the energy equal to work required to move an elementary charge e through a potential difference of 1 V

Units for electric field

Ch24-12/57

Electric potential:

a property of the electric field regardless of whether a charged object has been placed there:

-measured in J/C or V.

Electric potential energy:

is the potential energy of a charged object in an external electric field (i.e. potential energy of a system consisting of object and external electric field) :

-measured in joules.

Electric potential vs electric potential energy



Ch24-13/57

Apply a force to a particle to move it from i to f in an electric field. The applied force does work Wappwhile the electric field does work W on it.

By the work-kinetic energy theorem (Ch 7), the change in kinetic energy of the particle ∆K is:

If particle is stationary before and after, ∆K = 0and hence

i.e. the applied force exactly balances the workdone by the electric field as long as no change in KE

Work done by an applied force

Ch24-14/57

From before:

Hence:

i.e. if static before and after, the work done by the applied forces is equal to the change in the electric potential energy of the particle.

Recalling ∆V=-W/q and using the eqn at top we get

Wapp is the external work done to move a particle with charge q through a potential difference ∆V with no change in the particle’s KE.

Work done by an applied forceIf stationary before & after



Ch24-15/57

If we move the proton from i to f in the uniform electric field,

(a) do we do +ve or -ve work?

(b) does the proton move to a point of higher or lower potential?

Question

Ch24-16/57

If we move the proton from i to f in the uniform electric field,

(a) do we do +ve or -ve work?

(b) does the proton move to a point of higher or lower potential?

Question

Ans(a): our force does +ve work since our force and direction are aligned.

Ans(b): Wapp = q∆V, q and Wapp are +ve => ∆V is +ve =>the potential is higher. (which we got before (Ch24-6) considering ∆U = -W )



Ch24-17/57

Defn: A surface (real or imaginary) for which all points are at the same electric potential.

No net work is done on a charged particle by an electric field as long as the particle does not move offthe surface, or more generally, as long as its initial and final positions are on the equipotential surface.

This follows since ∆V=-W/q and hence W is 0 if ∆V=0.

Applies for any path!

Equipotential surfaces

Ch24-18/57

4 equipotential surfaces and four movements of particles.

I and II no net work is done because they both end on the same surface as they started on.

III and IV, the same net work is done because they both start and stop on the same surfaces (1 and 2)




Ch24-19/57

Theorem:Equipotential surfaces are | electric field lines and thus to E which is tangent to these lines.

Proof: Suppose E were not | the equipotential surface

=> it has a component lying along that surface

=> it would do work on a particle moving on the surface

But ∆V=-W/q so that work cannot be done as we move along a surface for which ∆V=0.

Hence E must be | to any equipotential surface.

QED


Ch24-20/57

In the above cases, we are only seeing the cross sections of the equipotential surfaces as dashed lines. The field lines are solid blue lines




Ch24-21/57

As qo goes along the path shown

and we know so that

and hence

Using ii above we get

If we choose potential at Vi as 0 =>

Conservative force => all paths same Note: independent of charge qo

Calculating the potential from the fieldTo calculate the potential difference between i & f, for any two points in an electric field:

i) calculate work done on a +ve test charge going from i to f

ii) apply

Ch24-22/57

Example: finding the potential differenceTwo points, i and f, in a uniform electric field E. They are on same field line (not shown), and are separated by d. Find Vf-Vi by moving +ve test charge qo along path shown.

Key idea: for a positive test charge

Here, since E constant and parallel to motion

=>

The -ve sign => potential decreases (the force has helped the +ve test charge move).



Ch24-23/57

Same situation but a different path. What is Vf-Vi ?

Key idea: Same approach as before.

path ic: since E | ds => Vc-Vi=0

path cf:

Example: finding the potential difference

Is there an easier way to solve this part of the problem?

Ch24-24/57

Key idea: The potential difference Vf-Vi from part (a)is -Ed. The potential difference between any two points is independent of the path taken.

Therefore the answer to part (b) is the same as (a) =-Ed.

Example: easier solution to part b



Ch24-25/57

Theorem: The potential due to a point charge q at any distance r is:

Proof: We take the potential at infinity as 0.

Consider moving qo from P to infinity:

Select the simplest path, radially away from q so that:

Write ds as dr, take Vf=0 and recall

Potential due to a point charge

QED after cancelling -ve signsand taking r=R since arbitrary

Ch24-26/57

We derived this for a +ve charge but the

derivation holds if it is -ve as well

Sign of V is the same as the sign of q.

+ve charge produces +ve potential

-ve charge produces -ve potential

Same formula applies anywhere outside or on surface of a spherically symmetric charge distribution since shell theorem tells us its electric field is the same as the point charge for which this eqn was derived.

Potential due to a point charge



Ch24-27/57

Superposition principle applies to electric potential.

For n charges qi, net potential is:

- algebraic sum,

- no vector addition required since V is a scaler quantity.

Potential due to a group of point charges

Ch24-28/57

Summary so far Ch 24

V +ve => electric field did -ve work 1 volt = 1 joule per coulomb

E must be | to any equipotential surface



Ch24-29/57

Three arrangements of two protons. Rank the arrangements in terms of the net electric potential at point P, largest first.

Question

Ans:

Which would be the arrangement with the weakest electric field at point P?Ans:

Which of a or b would have the largest electric field?Ans:

Ch24-30/57

Three arrangements of two protons. Rank the arrangements in terms of the net electric potential at point P, largest first.

Question

Ans: all three are the same since we sum 1/d and 1/D in all cases.

Which would be the arrangement with the weakest electric field at point P?Ans: (c) since forces are in opposite directions and hence cancel when add vectors whereas the other 2 add together. Which of a or b would have the largest electric field?Ans: (a) since the fields are aligned.



Ch24-31/57

What is the electric potential at point P if

q1 = +12 nC q2= -24 nC q3=+31 nC q4 = +17 nC

with d=1.3 m.

Key idea:

Potential is algebraic sum of individual potentials.

The distance in all cases is d/sqrt2 = 0.919 m.

Stuff in numbers to get V = 350 V

Sample problem

Net of 36 nC charge over 1.3 m produces a potential of 350 V

Ch24-32/57

12 electrons equally spaced around a circle of radius R. What are the electric potential and field at the centre C?

Key ideas:

Add the electric potential algebraically, and the potential from each electron is the same so

The electric field is added as a vector, and by symmetry they exactly cancel in pairs =>

Another sample problem



Ch24-33/57

What happens in (b) where the symmetry is partially gone?

Key ideas: Electric potential does not change because it adds algebraically and the distances and charges are unchanged.

The electric field now has only the y-components cancel but there will be a net electric field to the right, attracting a +ve test charge to the electrons.

Sample problem (cont)

Ch24-34/57

Now lets apply eqn for a group of point charges to an electric dipole

Natural dipoles (molecules) are very small so we assume r>>d (as before), in which case:

Potential due to an electric dipole

and

Unlike E case (z axis only) this applies at any angle

dipolep=qd,

-ve to +ve



Ch24-35/57

Ans:

Consider 3 points at r>>d around the dipole:

a is on the dipole axis above the dipole

b is on the dipole axis below the -ve charge

c is on the perpendicular bisector through the line joining the charges

Rank the points, greatest first(most +ve), in terms of potential at each point.

Question

a > c = 0 > b since cosθ = +1, 0, -1

Ch24-36/57

Why do we keep talking about dipoles?

-they are a good numerical example

-many molecules are polar and the results we develop apply to them (eg water)

-many nonpolar atoms and molecules have induced dipoles in an external electric field

Induced dipole moment

Nonpolar => +ve and -ve charges in the molecule are centered at the same point as in (a). When external electric field E is present, - +ve charge distribution shifts in direction of E- -ve charge shifts the other way => p exists in the direction of E- atom/molecule is said to be polarized by the field. When E removed, the polarization and p disappear.



Ch24-37/57

In general,

i) we choose a differential element of charge, dq,

ii) determine the differential element of potential, dV, at P due to dq,

iii) integrate over the entire charge distn.

Take V=0 at infinity and use

to write, for dV at point P from dq:

where r is distance from dq to P.

Thus total potential is:

Integral is over entire charge distn, but simpler than for E since no vector components need be considered.

Potential due to a continuous charge distn

Ch24-38/57

Non-conducting rod, uniform linear density λ, length L. Find the potential at P, a distance d from left end of rod.

dx is at a distance

Hence:

Integrate from x=0 to x=L using

Appendix E integral #17

Potential due to a line of charge



Ch24-39/57

Integrate from x=0 to x=L using AppE#17

Potential due to a line of charge

V is the sum of +ve dq, so should be +ve. Is the result always +ve? Yes, since term in [ ] is > 1What would happen if we integrated from L to 0 =>-ve sign ?We would have to remove it based on common sense!

Ch24-40/57

We calculated E on the axis of a non-conducting disk with uniform surface charge σ (C/m2). What is V(z) on the axis?

Consider dq in a ring, radius R’, width dR’.

Potential due to a charged disk

Note V > 0 for +ve charge, as it should be.



Ch24-41/57

Proof: Given V, draw equipotential surfaces in increments of dV. E must be | these surfaces (theorem Ch24-19).

A test charge qo moves between two surfaces in a step ds.Work done by field is -qo dV (in general W = -q dV from before) Also, in general W = E.ds = qo E(cosθ)ds and hence:

or

E cos θ is just Es, the component of E in the direction s. QED

To calculate V given E:

How do we calculate E given V?

Theorem: The component of E in any direction is given by:

Calculating E given V

Ch24-42/57

The component of E in any direction is the negative of the rate at which the electric potential changes in that direction.

Take s in turn to be the x, y and z axis:

A compact way of writing this is

Calculating E given V (cont)

For the simple situation in which E is uniform

where s is | equipotential surfaces ( 0 if parallel)

(said grad V)



Ch24-43/57

Question

3 pairs parallel plates with same separations. Between plates, E is uniform and perpendicular to the plates.

i) rank them, greatest first according to magnitude of E

ii) for which pair is the field rightward?

iii) an electron is released mid-way between plates 3, does it stay still, move right or left with constant velocity or acceleration

Ch24-44/57

Question

3 pairs parallel plates with same separations. Between plates, E is uniform and perpendicular to the plates.i) rank them, greatest first according to magnitude of Eii) for which pair is the field rightward?iii) an electron is released mid-way between plates 3, does it stay still, move right or left with constant velocity or accelerationAns i) 2 is greatest E=-∆V/∆s and ∆V = 200, 220 and 200 V

Ans ii) 1) V up to right => dV/dx > 0 => Ex < 0.

(2) is same, 3) reverse, i.e. Ex > 0 => case 3 E is to rightAns iii) E non-zero => force=> acceleration to left since q -ve



Ch24-45/57

Electric field from potential: charged disk

Given potential on the axis of a charged disk:

Determine the electric field on the axis.

By symmetry of the problem, E must be aligned along the axis for any value of z.

Thus we need only Ez (others 0 anyway):

Same as we developed before from Coulomb’s Law!

Ch24-46/57

Electric potential energy of a system of 2 point charges

Consider the energy associated with keeping two such charges in place (often previously we just asserted they were fixed).

To bring them together requires external work in this case. This energy is stored as potential energy by the system (assuming the kinetic energy does not change).

The electric potential energy of a system of fixed point charges is equal to the work done by an external agent to assemble that system, bringing each charge in from an infinite distance.

i.e. as before, with Ui=0



Ch24-47/57

As we bring in first charge, there is no work done. But for second we must overcome the work being done by field.

Recall and ∆V =

where q=q2 in 1st equation and q=q1 in 2nd so we have

Charges of same sign => U and work done are +ve

Charges of opposite sign => U and work done are -ve (to keep them stationary)

Note this section is all review.


Ch24-48/57

Again, and for first charge = 0.

For second charge, using the previous result

For the third charge, it is sum of work to bring it close to q1 and work to bring it close to q2.


So the total potential energy is the sum of all 3.

Note: result is independent of order because of symmetry.

Units: N m2/C2 (C2/m) = N m = J



Ch24-49/57

How much kinetic energy does an alpha particle (Z=2) need to get within 9.23 fm of a gold nucleus(Z=79)? Assume the approach is direct and ignore electrons.

alpha particle approaching gold nucleus

Assume gold nucleus is stationary since so much heavier.

Key idea: Mechanical energy of system is conserved, i.e., KEi + Ui = KEf +Uf. Initially U = 0 at a great distance but as alpha gets closer, KE is converted to U. At 9.23 fm, alpha particle stops and then `bounces’ back. So KEi = Uf since Ui=0 and KEf= 0.

Why ignore e- of gold atom?

Outside atom U=0 since neutral. Inside e-, E=0 so no effect. Also much lighter.

Recall 1eV = 1.602x10-19 J

Ch24-50/57

Potential of charged, isolated conductorTheorem:Excess charge placed on an isolated conductor will distribute itself on the surface of that conductor so that all points on the conductor, both on the surface and inside, come to the same potential. This also applies if there is an interior cavity with or without a net charge.

Proof: EE=0 within an isolated conductor and excess charge resides on the surface. In general:

EE=0 => integral = 0 => Vi = Vf for any pair of points within the conductor. QED

Alternatively, all points on the surface are at the same potential because E | surface of a conductor and hence E.ds = 0 along any path on surface.



Ch24-51/57

Plot of potential vs radius for isolated spherical conducting shell 1 m radius with a charge of 1µC. Outside shell

and constant inside by the theorem.

Figure b is the electric field which we could calculate directly or using Er = dV/dr.

We could also get V from E using

Potential of charged, isolated conductor

Note the strong fields for small charges.

Ch24-52/57

Isolated conductor in an external electric field

Placed in an external electric field, the charges inside a conductor arrange themselves on surface to exactly counterbalance external field. The external field also ends up being | conductor’s surface as proven before.

Once the charges were re-distributed, if you could magically make the conductor itself disappear, without moving the charges, then everything would stay fixed since there is no field acting on any of the charges, i.e.field is a property of the charges, not the conductor.



Ch24-53/57

Spark discharges: breakdown voltages

When a material is placed in an electric field, past a certain threshold, atoms or molecules are torn apart and the free electrons create an avalanche discharge. Below the threshold the field will not accelerate the electrons enough to ionize other molecules/atoms and create more free electrons.

In air, value is about 3 x 106 V/m = 3 MV/m = 3 kV/mm.

This value is typically 5 to 10 times larger in most solids.

Why is it more in solids?Ans: The electrons cannot travel so far before interacting in solids => do not accelerate to such high energies for a given field => higher breakdown field required

Ch24-54/57

The van de Graaff generatorConsider a conducting sphere with a small hole in it with a rubber belt arranged to `spray’ electrons onto the belt and remove them inside the sphere by a wire attached to the sphere.

What is the electric field inside the sphere?

Where does the charge go once it leaves the belt?What is electric field at surface of sphere?

What is the maximum electric field at the surface of the sphere?

What is the potential at the surface?



Ch24-55/57

The van de Graaff generatorConsider a conducting sphere with a small hole in it with a rubber belt arranged to `spray’ electrons onto the belt and remove them inside the sphere by a wire attached to the sphere.

What is the electric field inside the sphere?

Where does the charge go once it leaves the belt?What is electric field at surface of sphere?

What is the maximum electric field at the surface of the sphere?

What is the potential at the surface?

0: since inside a conductor

to the outer surface of sphere

3 MV/m before breakdown

Ch24-56/57

How much charge is on sphere just before breakdown?

So a 1 m sphere has 333µC max.

Smaller radius => lower potential before breakdown. Sharp points have high fields!

The van de Graaff generator(cont)

=> V = r E or E = V/r

What is the maximum voltage on a sphere of radius r m?

Emax = 3MV/m => V = 3r MV

So a 1 m radius sphere can reach 3 MV without breakdown.



Ch24-57/57

Review

This figure (from Prof Peter Watson’s web site) provides a useful overview.

Note also that potential is a continuous function

Ch24 potential

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