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PS05 F19 - sdsdsdsdsd
Physics Ii: Electromagnetism (PHYS 2213)
Cornell University
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Physics 2213 Solutions to Problem Set 5
Due 5:00 PM Friday, 10/11/
- High Voltage Spherical Capacitor with Dielectric
Let us consider a spherical capacitor in which the space between the conductors is first just vacuum, then is filled with an insulating material of dielectric constantκ. Assume that the total charges on the outer and inner conductors are +Qand−Q, respectively, and that these charges are uniformly distributed on the conductors’ surfaces atr=bandr=a, respectively. (a) Calculate the capacitance of the spherical capacitor with only vacuum in between the “plates”.Hint #1: find howQrelates to∆V. Hint #2: recall what we did in class during lecture 10.
Solution:
- We start by calculating the field between the two spherical surfaces.
Er(a < r < b) =
−Q
4 πǫ 0 r 2
.
Note the minus sign since the inner conductor is negatively charged. Here, we have ignored the edge effects.
- Integrating the field gives us the potential:
V(a < r < b) =−
∫r
b
Erdr=
∫r
b
Q
4 πǫ 0 r 2
dr=
Q
4 πǫ 0
(
1
b
−
1
r
)
.
The sign can be tricky. The important part is that the voltage near the negatively charged (inner) conductor is negative provided thatV(b) = 0. Both check out fine, so our expression looks legit.
- The absolute potential difference between the two surfaces is then:
∆V =V(b)−V(a) =
Q
4 πǫ 0
(
1
a
−
1
b
)
.
Note that ∆V has a positive value.
- SinceQ=C∆V: C=
4 πǫ 0 ( 1 a− 1 b)
(b) Now let’s assume that the capacitor has a dielectric of dielectricconstantκbetween the plates. What is the capacitance now? Show your work or explain your reasoning for full credit.
Solution:Note that dielectric materialreducesthe electric field by a factor ofκ compared to the case in vacuum:
Er(dielectric) =Er(vac)/κ.
It follows that the capacitance is increased by a factor ofκ.
C(dielectric) =κC(vac) = 4 πκǫ 0 (a 1 − 1 b)
.
This agrees with our expectation that dielectric material increases the capacitance.
(c) What is the magnitude of the bound chargeQ′, in terms ofQandκ, due to the polarization of the dielectric?
Solution:The net effect of the dielectric is toreducethe enclosed charge byκso that the field gets weakened by that constant. On the inner surface we have−Q+Q′=−Q/κ. On the outer surface we haveQ−Q′=Q/κ. Both lead toQ′=Q(1− 1 /κ) =−(−Q)(1− 1 /κ). The minus sign means the induced charges will be opposite of the electrode’s polarity.
(d) Express the magnitude of the electric field at a radius just slightly larger thana— that is, just outside the inner conductor atr=aand inside the dielectric — in terms of i. σ, κ, ǫ 0 ,and numerical constants only, where -σ ≡ Q/(4πa 2 ) is the magnitude of charge density on the conductor (σ <0 since the inner surface is negatively charged). ii. σ′, κ, ǫ 0 ,and numerical constants only, where σ′ ≡Q′/(4πa 2 ) is the magnitude of bound charge density on the surface of the insulator Note that when we previously foundE =σ/ǫ 0 just outside a conductor, there was no dielectric material.
Solution:The electric field at this location is given by
E=
−Q
4 πκǫ 0 a 2
.
In terms ofσ, we can writeE=−σ/(ǫ 0 κ). In terms ofσ′, we can writeE=−σ′/[ǫ 0 (κ−1)].
If we use the superposition of charge, the effective surface charge density isσeff= −σ+σ′=−σ/κ, which agrees with the previously found expressionE=σ/ǫ 0 just outside a conductorif all the chargesare considered.
(e) The energy density (per unit volume) in an electric field of magnitudeEin a dielectric material is u=
1
2
κǫ 0 E 2
Starting with an expression for the electric field between the conducting plates of a spher- ical capacitor, integrate the energy density to find the total energyUin the electric field:
U=
∫∫∫
u dV
Show that your result agrees with the general expression for energy stored in a capacitor,
U=
1
2
QV
- Connecting Spheres
INITIAL
FINAL
+Q
+Q
A
A
B
B
VA, i
VB, i
VB, f VA, f
2R
R
Consider two hollow spherical conductors, Conductor A with radius 2Rand Conductor B with radiusR, which are separated by a very large distanced≫R. Each conducting shell has non-negligiblethicknessa. Initially, each conductor has a charge +Q, and are not connected electrically. A wire is then connected between the two conductorsurfaces, allowing charge to possibly move between them. (a) Rank the surface charge density of each of the conductors, before (σA,i,σB,i) and after (σA,f,σB,f) the wire is connected, on both the inside and outside surfaces. (Youshould have 8 terms, one per surface (inner and outer) per shell (A and B) per case (before and after).)
Solution:Since there is no charge inside either sphere, the surface charge density is zero on theinnersurface before and after. The charge must distribute evenly on the outer surface, soσA,i= Q/(4π(2R) 2 ) =Q/(16πR 2 ), andσB,i= Q/(4πR 2 ). After the connection, the two conductors must have equal potential, soVA,f = VB,f, or kQA,f/(2R) =kQB,f/R. Charge conservation givesQA,f+QB,f= 2Q. Solving the two equations givesQA,f= 4Q/3 andQB,f= 2Q/3. SoσA,f= (4Q/3)/(4π(2R) 2 ) = Q/(12πR 2 ), andσB,f= (2Q/3)/(4πR 2 ) =Q/(6πR 2 ). Since 0< 1 / 16 < 1 / 12 < 1 / 6 < 1 /4, we have (σA,B,i,f)inner< σA,i,out< σA,f,out< σB,f,out< σB,i,out.
(b) Now rank the electric potential at each surface of each of the conductors, before (VA,i, VB,i) and after (VA,f,VB,f), in the same manner as in part (a). Argue your choice of ranking.
Solution:Since there is no field inside conductor, the voltage at inner and outer surface must equal, for both spheres before and after. Using the chargescalculated from part (a), we have:VA,i=kQ/(2R),VB,i=kQ/R, andVA,f=VB,f= 2Q/(3R). Since 1/ 2 < 2 / 3 <1, we have:VA,i< VA,f=VB,f< VB,i, andVinner=Vouterfor all 4 cases.
(c) Is the potential energy of the spheres greater before or after they are connected? Why is there a change in energy? Give a physical explanation for where the excess energy comes from or goes.
Solution:The initial energy between the two spheres isUi=kQ 2 /d. From part (a) we knowQA,f= 4Q/3 andQB,f= 2Q/3, soUf=k(4Q/3)(2Q/3)/d= (8/9)Ui. So the energy before is greater than after. The energy difference is usedto move the charge (an amount ofQ/3) from sphereB(a higher potential initially) to sphereA (a lower potential initially), and will dissipate as heat e. in the wire that connected the spheres. There is also actually energy associated with forming the spherical shell charge distributionitselffor each sphere. This is called the “self energy”, and is equal toU=QV 2. Calculation of the self energies is not required to obtain full mark.
(d) Finally, describe what will happen (and why) to the charge distribution and values of the electric potential if we fill the hollow conducting spheres with adielectric material once they are attached by the wire.
Solution:Since there is no charge enclosed by either sphere, the charge distribution remains unchanged, so do the potential values. Because there is no field inside the spheres, polarization of charges within the dielectric material doesnot occur.
- Image Charges
Figure 1: (a): Anonconductingsphere with positive charge density is placed above an infinite grounded conducting plane. You may assume the radius of the sphere is such that it does not touch the plane. (b) We model this as two nonconducting spheres of equal and opposite charge densitywithoutthe conducting plane between them.
When we look at charge distributions on conductors, we usually use what is known as the ‘Method of Images’. In this question, we will walk through a typical image problem–finding the surface charge distribution on an infinite conducting plane belowa charged sphere–and then ask you to apply this method to a slightly different scenario. (a) First, we consider two nonconducting charged spheres, centered a distance 2dapart, each with constant charge densityρ, but with opposite signs. Find the potential anywhere on thexy-plane.
WithEsurface=σ/ǫ 0 , we obtain:
σ=
− 2 da 3 ρ 3(x 2 +y 2 +d 2 ) 3 / 2
.
(d) Finally, use the method of images from your steps above to find the surface charge density if instead of a sphere we have an infinite nonconducting cylinder of charge densityρ=αr centered a distancedabove the infinite grounded conducting plane.
Solution:We know the field due to an infinite cylinder isE(r) = 2 πǫλ 0 rin the radial direction (at the least, we can prove this with Gauss’ Law). To calculateλwe consider a thin slice of the cylinder with thicknessdz. The charge of the slice isdQ=λdz= ρdV=
∫a 0 ρ(r)rdrdθdz
λ= 2πα
∫a
0
r 2 dr= 2παa 3 / 3.
Here, since the cylinder is along the y-axis, the distance from the top cylinder to any point in the xy-plane is: ~r r
=
xxˆ−dzˆ √ x 2 +d 2
(9)
and since they have opposite charge, theirExcancel and theirEzare equal on the xy-plane, as in part (c). The resultant field is therefore:
Ez= 2× −αda 3 3 ǫ 0 (x 2 +d 2 )
Note that the field is independent of the y coordinate since the cylinder is infinitely along the y direction. The minus sign indicates that the field is pointing in the−z direction. WithEsurface=σ/ǫ 0 , the surface charge density is
σ=− 2 αda 3 /3(x 2 +d 2 ).
- Capacitor Circuit
Given this circuit, where all capacitors have a capacitance ofCexceptC 2 , which has a capac-
itance of 43 C, find the following quantities after it has equilibrated:
(a) The charge onC 1
Solution:First find the equivalent capacitanceCeqseen by the power source. C 3 andC 4 are in parallel, soC 34 =C+C= 2C. C 34 andC 5 are in series, soC 345 = (2)(1)/(2 + 1)C= 23 C. C 345 andC 2 are in parallel, soC 2345 = 23 C+ 43 C= 2C. C 6 andC 7 are in series, soC 67 = (1)(1)/(1 + 1)C=C 2. C 67 andC 2345 are in series, soC 234567 = (2)( 12 )/(2 + 12 )C= 25 C. C 1 andC 234567 are in series, soCeq= (1)( 25 )/(1 + 25 )C= 27 C. Finally, sinceC 1 is in series with the source,Q 1 =V(Ceq) = 2 CV 7
(b) The charge onC 2
Solution:SinceC 1 andC 2345 are in series,Q 2345 =Q 1 = 2 CV 7. SinceC 345 andC 2 are in parallel, we have a charge divider: Q 2 =
(
C 2 C 2 +C 345
)
Q 2345 = 462 CV 7 = 4 CV 21
(c) Find the voltage overC 3
Solution:First find the charge onC 345. SinceC 345 andC 2 are in parallel, Q 345 =Q 2345 −Q 2 = 2 CV 7 − 4 CV 21 = 2 CV 21 (You can also show this using a charge divider). SinceC 34 andC 5 are in series,Q 34 =Q 345 = 2 CV 21. SinceC 3 andC 4 are in parallel,V 3 =V 34 =Q 34 /C 34 = 2 CV 21 /(2C) = 21 V.
PS05 F19 - sdsdsdsdsd
Course: Physics Ii: Electromagnetism (PHYS 2213)
University: Cornell University
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