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Unformatted text preview: 18.24 The three capacitors are connected in parallel, so Ceqzcl+02+03220nF +3.0,uF +4.0,uF 29.0,uF. 1825 The three capacitors are connected in series, so 1 1 1
__+ ,m 1 1 1 13 12 1
Ceq 01 C—ZT03m2.GpF+3.OpF+4.OpF 12ppﬂ>Ceq=§prxas32as 18.35 etc that the two capacitors are connected in parallel with the independent voltage source, and so they
both have the same potential difference across them, 15 V. 3)} The magnitude of the charge on each capacitor is found from the potentiai difference and its capacitance, J9. a
Ca m => go: Oil/i. ; For the 4.0 ,LLF capacitor, Q : (4.0 x 105 r)(15 V) 2 6.0 X10_0 0 m 60,110.
For the 6.0 ,LLF capacitor, Q=(6.0 x 106 F)(15V) : 9.0 x 105 c = gone. (i) The two capacitors are connected in parallel, so the equivaient capacitance is their sum 05,, z 4.0 as + 6.0 ,ur a 10.0 F. 18.36 Let C and C" be the capacitances. When they are connected in parallel the equivalent capacitance is 12 pl“, so
12 pF 2 C + C'.
When connected in series, the equivalent capacitance is one third of, say, 0, so
1 CC, 2 . r r r
EC:C+C, m C “FCC 2300 :;> 0—20. Use this resuit in first equation, and ' 18.43 Let n be the number of capacitors in the bank. Each stores a potential energj,r of
1 2
50V ,
so we want 1 _ 2 = 
MN ={n)%CV2 => 10.1 = (n)—2{1(}U><1(} 6F)(100V) :> n 20 18.44 't de
a) The electricai force on the charge and the force of the professor on the charge are of eqiaalogiizgiihe
and opposite direction, so the total force on the charge is zero and 1t Inoves at constant spiethe work done
Plates of the capacitor. Hence, the work done by the electrical force Will be the negative 0
by the professor. So the work done by the electricai force is ~2.0 J  . . . ' n ' lenere 0f the
bl The work done by the electrical force is related to the change in the electrical Potentia’ ﬂy .
charge by the equation  2.0.1 3
— 2 — . 1 V.
Wanna: ~APE ==> —2.0J = —{qAV) => AV_ #20 X 10_3 G 10x 0 . ' ’ l e,
Hence, the potential difference between the plates of the capac1tor, taken as the high minus the iow vs. 11 _
is 1.0 x 103 V. 18.46 a) The magnitude of the electric ﬁeid at the surface of a uniform Sphericai charge distribution of radiue R
is
1
E : lQE RE 2' : 47TEORZE (6.37 x105 to H100 N/C) 5
3* WFWﬂM “0 G Since the direction of the ﬁeid is tcward the surface, i.e., toward the charge, the charge must be negative,
which means that Q = —4.5} x 105 C. b) The energy density is 1 1 4
energy density : —2—EQE2 : i 37:0 (1001\1/(3)2
47T(9.00 X 109 Nmg/Cz) E2 2% —_~ 4.42 ><10‘8 J/m3. /, w." _.,.,...w._, :uuu cwuiuuiciauy avaziaoie! £2.58 Let C’ be the capacitance of the system with me dielectric and C" be the capacitance with a dielectric.
en 0’2n0 ==>100aF=26CI => C=3.8,uF. Therefore, the effect of the dielectric is to raise the capacitance frem 3.8 ,uF to 100 ,LLF, that iS, by EL faCtO‘f
of almost 25 (1e, by the dielectric constantit). ...
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This note was uploaded on 10/01/2009 for the course PHYS 102 taught by Professor Henry during the Spring '09 term at Johns Hopkins.
 Spring '09
 HENRY
 Physics

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