• Share
  • Email
  • Embed
  • Like
  • Save
  • Private Content
Ch24 potential
 

Ch24 potential

on

  • 224 views

 

Statistics

Views

Total Views
224
Views on SlideShare
224
Embed Views
0

Actions

Likes
0
Downloads
4
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Ch24 potential Ch24 potential Document Transcript

    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons HRW628-645 Ch 24: Electric Potential If a force is conservative => we can associate a potential energy with it. => principle of conservation of mechanical energy applies in closed systems involving the force. Review ch 7 and 8. -change in potential energy: W is the work done on the particles by the applied force. -change in kinetic energy same direction -work by a constant force =>+ve work Ch24-1/57 Electric Potential Energy 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 (Force does +ve work => potential decreases) The work done by a conservative force is path independent (assumes rest of the system doesn’t change) Potential energy requires a reference configuration and a reference potential energy be set/defined. Usually take reference condition as all particles infinitely separated and define this to have 0 potential energy. Ch24-2/57Material, 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 CopyrightProtection. 1
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Electric potential energy: Work done by a field 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. Ch24-3/57 Electric potential energy: work done by a field Earth’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? 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. Ch24-4/57Material, 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 CopyrightProtection. 2
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question 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? Ch24-5/57 Question 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? Ans: a) W = qEd cos180o = -qEd and q > 0 => W is -ve (i.e. external force has to work to do it). Ans b) >0 so the potential increases Ch24-6/57Material, 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 CopyrightProtection. 3
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons The sign of work 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 Ch24-7/57 Electric potential 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 Ch24-8/57Material, 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 CopyrightProtection. 4
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Electric potential difference Potential difference between two points is the negative of 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. Ch24-9/57 Electric potential difference The SI unit is J/C which is given the special name volt, V (cap since named after a person). Count Alessandro 1 volt = 1 joule per coulomb Volta Italian 1745-1827 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 Ch24-10/57Material, 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 CopyrightProtection. 5
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Units for electric field 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 Ch24-11/57 Electric potential vs electric potential energy 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. Ch24-12/57Material, 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 CopyrightProtection. 6
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Work done by an applied force Apply a force to a particle to move it from i to f in an electric field. The applied force does work Wapp while 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 = 0 and hence i.e. the applied force exactly balances the work done by the electric field as long as no change in KE Ch24-13/57 Work done by an applied force If stationary before & after 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. Ch24-14/57Material, 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 CopyrightProtection. 7
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question 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? Ch24-15/57 Question 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? 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 ) Ch24-16/57Material, 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 CopyrightProtection. 8
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Equipotential surfaces 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 off the 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! Ch24-17/57 Equipotential surfaces 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) Ch24-18/57Material, 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 CopyrightProtection. 9
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Equipotential surfaces 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 Ch24-19/57 Equipotential surfaces 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 Ch24-20/57Material, 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 CopyrightProtection. 10
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Calculating the potential from the field To 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 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 Ch24-21/57 Example: finding the potential difference Two 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). Ch24-22/57Material, 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 CopyrightProtection. 11
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Example: finding the potential difference 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: Is there an easier way to solve this part of the problem? Ch24-23/57 Example: easier solution to part b 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. Ch24-24/57Material, 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 CopyrightProtection. 12
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Potential due to a point charge 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 QED after cancelling -ve signs and taking r=R since arbitrary Ch24-25/57 Potential due to a point charge 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. Ch24-26/57Material, 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 CopyrightProtection. 13
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Potential due to a group of point charges 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. Ch24-27/57 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 Ch24-28/57Material, 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 CopyrightProtection. 14
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question Three arrangements of two protons. Rank the arrangements in terms of the net electric potential at point P, largest first. 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: Ch24-29/57 Question Three arrangements of two protons. Rank the arrangements in terms of the net electric potential at point P, largest first. 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. Ch24-30/57Material, 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 CopyrightProtection. 15
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem 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 Net of 36 nC charge over 1.3 m produces a potential of 350 V Ch24-31/57 Another sample problem 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 => Ch24-32/57Material, 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 CopyrightProtection. 16
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem (cont) 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. Ch24-33/57 Potential due to an electric dipole Now lets apply eqn for a group of point charges to an electric dipole dipole p=qd, Natural dipoles (molecules) are very small so -ve to +ve we assume r>>d (as before), in which case: and Unlike E case (z axis only) this applies at any angle Ch24-34/57Material, 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 CopyrightProtection. 17
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question 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. Ans: a > c=0 > b since cosθ = +1, 0, -1 Ch24-35/57 Induced dipole moment 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 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. Ch24-36/57Material, 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 CopyrightProtection. 18
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Potential due to a continuous charge distn 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. Ch24-37/57 Potential due to a line of charge 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 Ch24-38/57Material, 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 CopyrightProtection. 19
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Potential due to a line of charge Integrate from x=0 to x=L using AppE#17 V is the sum of +ve dq, so should be +ve. Is the result always +ve? Yes, since term in [ ] is > 1 What would happen if we integrated from L to 0 =>-ve sign ? We would have to remove it based on common sense! Ch24-39/57 Potential due to a charged disk 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’. Note V > 0 for +ve charge, as it should be. Ch24-40/57Material, 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 CopyrightProtection. 20
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Calculating E given V To calculate V given E: How do we calculate E given V? Theorem: The component of E in any direction is given by: 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 Ch24-41/57 Calculating E given V (cont) 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 (said grad V) For the simple situation in which E is uniform where s is | equipotential surfaces ( 0 if parallel) Ch24-42/57Material, 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 CopyrightProtection. 21
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons 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 Ch24-43/57 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 Ans 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 right Ans iii) E non-zero => force=> acceleration to left since q -ve Ch24-44/57Material, 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 CopyrightProtection. 22
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons 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! Ch24-45/57 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 Ch24-46/57Material, 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 CopyrightProtection. 23
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Electric potential energy of a system of 2 point charges 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. Ch24-47/57 Electric potential energy of a system of 3 point charges 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 Ch24-48/57Material, 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 CopyrightProtection. 24
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons alpha particle approaching gold nucleus 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. 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 Ch24-49/57 Potential of charged, isolated conductor Theorem: 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: E=0 within an isolated conductor and excess charge resides on the surface. In general: E=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. Ch24-50/57Material, 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 CopyrightProtection. 25
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Potential of charged, isolated conductor 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 Note the strong fields for small charges. Ch24-51/57 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. Ch24-52/57Material, 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 CopyrightProtection. 26
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons 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 Ch24-53/57 The van de Graaff generator Consider 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? Ch24-54/57Material, 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 CopyrightProtection. 27
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons The van de Graaff generator Consider 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? 0: since inside a conductor Where does the charge go once it leaves the belt? to the outer surface of sphere What is electric field at surface of sphere? What is the maximum electric field at the surface of the sphere? 3 MV/m before breakdown What is the potential at the surface? Ch24-55/57 The van de Graaff generator(cont) => V=rE 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. 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! Ch24-56/57Material, 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 CopyrightProtection. 28
    • Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Review This figure (from Prof Peter Watson’s web site) provides a useful overview. Note also that potential is a continuous function Ch24-57/57Material, 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 CopyrightProtection. 29