Physics: Electromagnetism Spring PROBLEM SET 6 Solutions

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1 Physics: Electomagnetism Sping 7 Physics: Electomagnetism Sping 7 PROBEM SET 6 Solutions Electostatic Enegy Basics: Wolfson and Pasachoff h 6 Poblem 7 p 679 Thee ae si diffeent pais of equal chages and the sepaation of any pai is a Thus W = pais kqq i j / a= 6 kq / a 3 apacito Basics -- ylindical Symmety: Wolfson and Pasachoff h 6 Poblem 35 p 68 The capacitance of ai-filled ( κ = ) cylindical capacito was found in Eample 6-5 in you tetbook and in the class eample Utilizing that epession we obtain: πε π (885 pf/m)( m) = = = 55 pf ln( b/ a) ln( / 8) 4 apacito Basics -- Spheical Symmety: Wolfson and Pasachoff h 6 Poblem 36 p 68 et s assume that the oute shell with adius b is caying a chage +Q and the inne sphee of adius b is caying a chage -Q Then the field outside the two sphees is zeo and the field between the sphees is just kq/ It is clea then that the potential between the two sphees is a b a a kq kq kq kq V = d = d d kq = + = b b a b Theefoe the capacitance of such a capacito is Q Q ab ab = = = V kq b a k b a ( ) ( ) 5 Enegy Stoed in apacitos: Wolfson and Pasachoff h 6 Poblem 47 p 68 a) The chage on the plates emains the same and so does the electic field ( E = σ / ε ) in the gaps between eithe plate and the slab Howeve the sepaation (ie the thickness of the field egion) between the plates is educed to 4% of its oiginal value d = d+ d = 4d theefoe the capacitance is inceased = ε A/ d = ε A/ 4d = 5 (The equations V = El and = Q/ V lead to the same esult) In fact the configuation behaves like a seies combination of two paallel plate capacitos / = + = ( d / ε A) + ( d / ε A) = ( d + d ) / ε A= 4 d/ ε A= / 5 b) When the chage is constant (no connections to anything isolates the system) the enegy stoed is invesely popotional to the capacitance U = Q / Thus U = Q / = Q /(5 ) = 4 U o the enegy deceases to 4% of its oiginal value (With the slab inseted thee is less field egion and less enegy stoed While the slab is being inseted wok is done by electical foces to conseve enegy)

2 Physics: Electomagnetism Sping 7 6 onnecting apacitos: Wolfson and Pasachoff h 6 Poblem 5 p 68 (i) is in seies with the paallel combination of and 3 Thus ( + 3) ( µ F) ( + ) = = = µ F ( + + 3) (+ + ) (ii) The net chage on the entie combination is Q= = ( µ F)( V) = µ Since is in seies with the capacitos in paallel Q= µ = Q = Q + Q3 Moeove fo the paallel capacitos V = Q / = V3 = Q3/ 3 so Q3/ Q = 3/ = Thus Q = (/ 3) Q = 4 µ and Q3 = (/3) Q = 8 µ (In geneal fo two capacitos in paallel Q = Q/( + 3) etc) (iii) Equation 6-5 in you tetbook and the ones deived in class applied to each capacito give = / = / F = 6 V V Q µ µ And V = V3 = 4 V 7 Biophysics!!! Wolfson and Pasachoff h 6 Poblem 7 p 68 If we assume that the inne and oute sufaces of the membane act like a paallel plate capacito with the space between the plates filled with mateial of dielectic constant κ = 3 then the capacitance pe unit aea is κε = A d Thus 3(885 pf/m) d = 7 nm ( F/cm ) = µ 8 Moe Biophysics!!! A neuon o neve cell has a long signal-caying aon (which fom spine to finges can be moe than a mete in length) The aon membane is usually positive on the outside and negative on the inside The dielectic constant of the membane has been measued to be about 7 a) Given that the membane wall is a mee 6 µm thick and that the aon adius is 5 mm detemine its capacitance pe unit aea (Hint: the aon is a tube Because the wall thickness is much smalle than both the inne and oute diametes of the tube you can teat it as a paallel-plate capacito)

3 Physics: Electomagnetism Sping 7 The aon can be teated as a paallel plate capacito whose capacitance is ε = κε Then the capacitance of the aon pe unit aea is A ε d 7 F/m 6 m 885 F/m = κ = = 9 = ε A d whee b) Detemine the suface chage density of the aon In the esting state when thee is no signal being tansmitted the potential diffeence acoss the membane is about 7 mv Using the definition of capacitance = Q/ V we can find the suface chage density of the aon to be Q 3 4 σ = = = ( F/m )( 7 V) = 7 /m A A whee we have used the esult of pat (a) fo the capacitance of the aon pe unit aea c) When a pulse is popagated down the aon thee is a shift of ions acoss a segment of membane that causes a evesal of polaity and an oveall change in potential of about mv This voltage spike (and the associated epolaization of the membane) popagates along fom one egion to the net and constitutes the signal just as a flame popagates down a length of fuse Afte a few milliseconds this so-called action potential pulse passes and given point of the aon which then etuns to its esting potential (~7 mv) How much enegy is equied to echage - m length of aon in the wake of a pulse so that it will be eady to tansmit the net pulse? Fo a -m long aon the capacitance is = ( F/m ) ( 5 6 m)( m) 3 7 F A π = π = The enegy needed to echage the aon is ( 3 7 F 3 )( V ) 6 9 U = = = J = 6 nj 9 Enegy Stoed in an Isolated and onnected apacitos a) Wolfson and Pasachoff h 6 Poblem 8 p 683 a Ignoing finging fields the electic field between the plates is unifom V E = d whee V V since the dielectic is inseted and E depends on On the left side of the capacito on Figue 6-36 whee the slab has penetated σ l E = κ ε and on the ight of the capacito E σ = ε whee σ l and σ ae the chage densities on the left and ight sides espectively Thus σ l = κεe and σ = ε E and the chage can be witten in tems of geometical vaiables supeposed on Figue below as 3

4 Physics: Electomagnetism Sping 7 V q= σ w+ σw( ) = εew( κ+ ) = ε w( κ+ ) d Then the capacitance can be witten as q ( κ + ) = = () V ε A whee = and A= w Although the question specifies = fo which value the d capacitance is = ( κ + ) we give as a function of in () because we will need to diffeentiate with espect to in pat (c) b When the battey is disconnected the capacito is isolated and the chage on it is a constant q= q The stoed enegy then is q q U U = = = ( κ+ ) ( κ+ ) whee U q = = Fo = the enegy is U = ( κ + ) c The foce on a pat of an isolated system is elated to the potential enegy of the system by F = U () The foce due to the capacito on the slab acts in one dimension along -ais and theefoe is F HG du d U F = d = d κ + I = KJ U ( κ ) ( κ + ) in the diection of inceasing (so as to pull the slab into the capacito) Fo = the magnitude of the foce is ( κ ) F = ( κ + ) It tuns out that if we ewite the foce fo any value of in tems of the voltage fo that using ( κ + ) q = = = the epession can be used in the succeeding poblem Thus F κ V ( κ ) ( ) ( κ ) = = = ( κ+ ) b) Wolfson and Pasachoff h 6 Poblem 8 p 683 The capacitance depends on the configuation and electical popeties of the plates and insulating 4

5 Physics: Electomagnetism Sping 7 mateials not on the etenal connections so ( κ + ) = as in the peceding poblem (i) If the capacito emains connected to a battey the voltage is constant V = V and Fo = U = = ( κ + ) U = 4 ( κ + ) (This is diffeent fom the peceding poblem because the battey does wok) i When the capacito is connected to a battey Equation () above fo the foce does not apply since the chage on plates is a function of both potential enegy and continuously changing chage (The foce in this case is deived in moe advanced tets) Howeve fo paticula values of chage and voltage on the capacito the foce on the slab consideed hee is the same egadless of the etenal connections In the peceding poblem we found that ( κ ) F = whee V was the paticula voltage (and because of the special fom of the capacitance ( ) the paticula chage q did not appea -- necessay condition fo this fomula to be applicable in this poblem) Since V = V in this poblem y F ( κ ) = σ w σ An Engineeing Poblem A paallel-plate capacito has the space between the plates filled with two slabs of dielectic one with constant Κ and one with constant Κ as shown in the figue below Each slab has thickness d / whee d is the plate sepaation Show that the capacitance is 5

6 Physics: Electomagnetism Sping 7 ε A κκ = d κ+ κ K d/ K d/ The capacitance of a capacito filled with two slabs of dielectics of dielectic constants κ and κ is equivalent to that of two capacitos in seies with the above two dielectic constants espectively: εa εa d d d d d εa εa εa κ κ = + = + = + Then simplifying we obtain ε A κκ = d κ + κ 6