1. Electrical energy & Capacitance PHY232 Remco Zegers zegers@nscl.msu.edu Room W109 – cyclotron building http://www.nscl.msu.edu/~zegers/phy232.html

2. PHY232 - Remco Zegers - Electrical Energy & Capacitance 2 work…  previously… A force is conservative if the work done on an object when moving from A to B does not depend on the path followed. Consequently, work was defined as: W=PE i -PE f =-  PE  this was derived for a gravitational force, but as we saw in the previous chapter, gravitational and Coulomb forces are very similar: F g =Gm 1 m 2 /r 12 2 with G=6.67x10 -11 Nm 2 /kg 2 F e =k e q 1 q 2 /r 12 2 with k e =8.99x10 9 Nm 2 /C 2 Hence: The Coulomb force is a conservative force

3. PHY232 - Remco Zegers - Electrical Energy & Capacitance 3 work & potential energy  consider a charge +q moving in an E field from A to B over a distance D. We can ignore gravity (why?)  What is the work done by the field?  What is the change in PE?  If initially at rest, what is its speed at B?  W AB =Fdcos  with  the angle between F and direction of movement, so  W AB =Fd  W AB =qEd (since F=qE)  work done BY the field ON the charge (W is positive)   PE=-W AB =-qEd : negative, so the potential energy has decreased  Conservation of energy:   PE+  KE=0   KE=1/2m(v f 2 -v i 2 )  1/2mv f 2 =qEd  v=  (2qEd/m)

4. PHY232 - Remco Zegers - Electrical Energy & Capacitance 4 work & potential energy II  Consider the same situation for a charge of –q.  Can it move from A to B without an external force being applied, assuming the charge is initially (A) and finally (B) at rest?  W AB =-qEd ; negative, so work must be done by the charge. This can only happen if an external force is applied  Note: if the charge had an initial velocity the energy could come from the kinetic energy (I.e. it would slow down)  If the charge is at rest at A and B: external work done: -qEd  If the charge has final velocity v then external work done: W=1/2mv f 2 +|q|Ed

5. PHY232 - Remco Zegers - Electrical Energy & Capacitance 5 Conclusion  In the absence of external forces, a positive charge placed in an electric field will move along the field lines (from + to -) to reduce the potential energy  In the absence of external forces, a negative charge placed in an electric field will move along the field lines (from - to +) to reduce the potential energy +++++++++++++ --------------------

6. PHY232 - Remco Zegers - Electrical Energy & Capacitance 6 question  a positive charge initially at rest at P moves to Q. Will it follow the shortest route (as indicated by the dashed arrow) a) yes b) no  Will the change in potential energy of the charge at Q be different depending on which path is taken from P to Q? a) yes b) no c) depends on whether the velocity of the charge at Q is different depending on the path d) depends on the external forces applied P Q

7. PHY232 - Remco Zegers - Electrical Energy & Capacitance 7 question  a negatively charged (-1  C) mass of 1 g is shot diagonally in an electric field created by a negatively charge plate (E=100 N/C). It starts at 2 m distance from the plate and stops 1 m from the plate, before turning back. What was the initial velocity in the direction along the field lines? ---------------------- 2m 1m X Y

8. PHY232 - Remco Zegers - Electrical Energy & Capacitance 8 answer

9. PHY232 - Remco Zegers - Electrical Energy & Capacitance 9 Electrical potential  The change in electrical potential energy of a particle of charge Q in a field with strength E over a distance d depends on the charge of the particle:  PE=-QEd  For convenience, it is useful to define the difference in electrical potential between two points (  V), that is independent of the charge that is moving:  V=  PE/Q=-|E|d  The electrical potential difference has units [J/C] which is usually referred to as Volt ([V]). It is a scalar  Since  V= -Ed, so E= -  V/d the units of E ([N/C] before) can also be given as [V/m]. They are equivalent, but [V/m] is more often used.

10. PHY232 - Remco Zegers - Electrical Energy & Capacitance 10 Electric potential due to a single charge  the potential at a distance r away from a charge +q is the work done in bringing a charge of 1 C from infinity (V=0) to the point r: V=k e q/r  If the charge that is creating the potential is negative (-q) then V=- k e q/r  If the field is created by more than one charge, then the superposition principle can be used to calculate the potential at any point + r V 1C V=k e q/r

11. PHY232 - Remco Zegers - Electrical Energy & Capacitance 11 example +1C -2 C r 1 m a)what is the electric field at a distance r? b)what is the electric potential at a distance r? 1 2

12. PHY232 - Remco Zegers - Electrical Energy & Capacitance 12 question  a proton is moving in the direction of the electric field. During this process, the potential energy …… and its electric potential …… a)increases, decreases b)decreases, increases c)increases, increases d)decreases, decreases + + -

13. PHY232 - Remco Zegers - Electrical Energy & Capacitance 13 example  a particle (q 1 ) with a charge of +4.5  C is fixed in space. From a distance of 3.70 cm, a particle (q 2 ) of mass 6.9 g and charge –3.10  C is fired with initial velocity of 60 m/s towards to fixed charge. What is its velocity when it is 1 cm away from q 1 ?

14. PHY232 - Remco Zegers - Electrical Energy & Capacitance 14 question +1 -2 +1 A 1)the electric potential at A is a)zero b)non-zero 2)the electric field at B is a)zero b)non-zero 3)a + particle at A would a)move b)not move +1 B 1)the electric potential at B is a)zero b)non-zero 2)the electric field at B is a)zero b)non-zero 3)a + particle at A would a)move b)not move +1

15. PHY232 - Remco Zegers - Electrical Energy & Capacitance 15 equipotential surfaces compare with a map

16. PHY232 - Remco Zegers - Electrical Energy & Capacitance 16 A capacitor  is a device to create a constant electric field. The potential difference V=Ed  is a device to store charge (+ and -) in electrical circuits.  the charge stored Q is proportional to the potential difference V: Q=CV  C is the capacitance, units C/V or Farad (F)  very often C is given in terms of  F (10 -6 F), nF (10 -9 F) pF (10 -12 F)  Other shapes exist, but for a parallel plate capacitor: C=  0 A/d where  0 =8.85x10 -12 F/m and A the area of the plates +++++++++++++ -------------------- +Q -Q d symbol for capacitor when used in electric circuit

17. PHY232 - Remco Zegers - Electrical Energy & Capacitance 17 electric circuits: batteries  The battery does work (e.g. using chemical energy) to move positive charge from the – terminal to the + terminal. Chemical energy is transformed into electrical potential energy.  Once at the + terminal, the charge can move through an external circuit to do work transforming electrical potential energy into other forms Symbol used in electric circuits: +-

18. PHY232 - Remco Zegers - Electrical Energy & Capacitance 18 Our first circuit  The battery will transport charge from one plate to the other until the voltage produced by the charge build-up is equal to the battery charge  example: a 12V battery is connected to a capacitor of 10 nF. How much charge is stored?  answer Q=CV=10x10 -9 x 12V=120 nC  if the battery is replaced by a 300 V battery, and the capacitor is 2000  F, how much charge is stored?  answer Q=CV=2000x10 -6 x 300V=0.6C  We will see later that this corresponds to 0.5CV 2 =90 J of energy, which is the same as a 1 kg ball moving at a velocity of 13.4 m/s 12V 10nF

19. PHY232 - Remco Zegers - Electrical Energy & Capacitance 19 capacitors in parallel  We can replace C 1 and C 2 with one equivalent capacitor: Q 1 =C 1 V & Q 2 =C 2 V is replaced by: Q=C eq V since Q=Q 1 +Q 2, C 1 V+C 2 V=C eq V so:  C eq =C 1 +C 2  This holds for any combination of parallel placed capacitances C eq =C 1 +C 2 +C 3 +…  The equivalent capacitance is larger than each of the components 12V C 1 =10nF C 2 =10nF At the points the potential is fixed to one value, say 12V at A and 0 V at B This means that if the capacitances C 1 and C 2 are equal they must have the same charge stored and the total charge stored is Q=Q 1 +Q 2. A B

20. PHY232 - Remco Zegers - Electrical Energy & Capacitance 20 capacitors in series  we can again replace C 1 and C 2 with one equivalent capacitor but now we start from: V=V 1 +V 2 so, V=Q/C 1 +Q/C 2 =Q/C eq and thus: 1/C eq =1/C 1 + 1/C 2  This holds for any combination of in series placed capacitances 1/C eq =1/C 1 +1/C 2 +1/C 3 +…  The equivalent capacitor is smaller than each of the components 12V C 1 =10nF C 2 =10nF A B The voltage drop of 12V is over both capacitors. V=V 1 +V 2 The two plates enclosed in are not connected to the battery and must be neutral on average. Therefore the charge stored in C 1 and C 2 are the same

21. PHY232 - Remco Zegers - Electrical Energy & Capacitance 21 question  Given three capacitors of 1 nF, an capacitor can be constructed that has minimally a capacitance of: a)1/3 nF b)1 nF c)1.5 nF d)3 nF

22. PHY232 - Remco Zegers - Electrical Energy & Capacitance 22 a more general case: what is the equivalent C  step 1: C 4 and C 5 and C 6 are in parallel. They can be replaced by once equivalent C 456 =C 4 +C 5 +C 6 12V C1C1 C2C2 C3C3 C4C4 C5C5 C6C6 STRATEGY: replace subgroups of capacitors, starting at the smallest level and slowly building up.

23. PHY232 - Remco Zegers - Electrical Energy & Capacitance 23 step II  C 3 and C 456 are in series. Replace with equivalent C: 1/C 3456 =1/C 3 +1/C 456 so C 3456 =C 3 C 456 /(C 3 +C 456 )  C 1 and C 2 are in series. Replace with equivalent C: 1/C 12 =1/C 1 +1/C 2 so C 12 =C 1 C 2 /(C 1 +C 2 ) 12V C1C1 C2C2 C3C3 C 456

24. PHY232 - Remco Zegers - Electrical Energy & Capacitance 24 step III  C 12 and C 3456 are in parallel, replace by equivalent C of C 123456 =C 12 +C 3456 12V C 3456 C 12 12V C 123456

25. PHY232 - Remco Zegers - Electrical Energy & Capacitance 25 problem 12V C1C1 C2C2 C3C3 C4C4 C5C5 A B C 1 =10nF C 2 =20nF C 3 =10nF C 4 =10nF C 5 =20nF What is V ab ?

26. PHY232 - Remco Zegers - Electrical Energy & Capacitance 26 energy stored in a capacitor  the work done transferring a small amount  Q from – to + takes an amount of work equal to  W=V  Q  At the same time, V is increased, since V=(Q+  Q/C)  The total work done when moving charge Q starting at V=0 equals: W=1/2QV=1/2(CV)V=1/2CV 2  Therefore, the amount of energy stored in a capacitor equals: E C =1/2CV 2 +++++++++++++ -------------------- +Q -Q V Q V QQ

27. PHY232 - Remco Zegers - Electrical Energy & Capacitance 27 example  A parallel-plate capacitor is constructed with plate area of 0.40 m 2 and a plate separation of 0.1mm. How much energy is stored when it is charged to a potential difference of 12V?

28. PHY232 - Remco Zegers - Electrical Energy & Capacitance 28 capacitors II  the charge density of one of the plates is defined as:  =Q/A  The equation C=  0 A/d assumes the area between the plates is in vacuum (free space)  If the space is replaced by an insulating material, the constant  0 must be replaced by  0 where  (kappa) is the dielectric constant for that material, relative to vacuum  Therefore: C=  0 A/d +++++++++++++ -------------------- +Q -Q d A material  vacuum1.00000 air1.00059 glass5.6 paper3.7 water80

29. PHY232 - Remco Zegers - Electrical Energy & Capacitance 29 why does inserting a plate matter? molecules, such as water, can/is be polarized when placed in an E-field, the orient themselves along the field lines; the negative plates attracts the positive side of the molecules near to positive plate, net negative charge is collected; near the negative plate, net positive charge is collected. If no battery is connected, the initial potential difference V between the plates will drop to V/ . If a battery was connected, more charge can be added, increasing the capacitance from C to C 

30. PHY232 - Remco Zegers - Electrical Energy & Capacitance 30 problem  An amount of 10 J is stored in a parallel plate capacitor with C=10nF. Then the plates are disconnected from the battery and a plate of material is inserted between the plates. A voltage drop of 1000 V is recorded. What is the dielectric constant of the material?

31. PHY232 - Remco Zegers - Electrical Energy & Capacitance 31 problem  An ideal parallel plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected and the separation between the plates is increased in such a way that no charge leaks off. The energy stored in the capacitor has a)increased b)decreased c)not changed d)become zero

32. PHY232 - Remco Zegers - Electrical Energy & Capacitance 32 Remember  Electric force acting on object 1 (or 2): F=k e q 1 q 2 /r 12 2  Electric field due to object 1 at a distance r: E=k e q 1 /r 2  Electric potential at a distance r away from a charge q 1 : V=k e q 1 /r