(a) Calculate the resulting charges Q L and Q R on the left and right capacitorr
ID: 1737010 • Letter: #
Question
(a) Calculate the resulting charges QL andQR on the left and right capacitorrespectively.
QL = C
QR = C
(b) Calculate the total energy stored in the capacitornetwork.
Utot = J
(c) Calculate the new values of the charges on eachcapacitor.
QL' = C
QR' = C
(d) Calculate the subsequent voltage acrossCR.
VR' = V
(e) What is the total energy stored in the circuit in its finalconfiguration?
Utot' = J
(f) Compare the total energy stored in the capacitor networkbefore and after the dielectric is inserted, and from thisdetermine whether the dielectric is pulled intoCR or must be pushed intoCR.
Answer 1 if pulled, 2 if pushed:
An E = 20 V battery isused to charge two identical capacitors CL =CR = 4 µF, as shown in this diagram.(a) Calculate the resulting charges QL andQR on the left and right capacitorrespectively.
Explanation / Answer
(a)The capacitors CL and CR areconnected in parallel.Therefore,the potential difference across thetwo capacitor is same.The resulting charges QL and QR on the leftand right capacitor respectively are QL= CL* E = 4 * 10-6 * 20 =80 * 10-6 C and QR= CR* E = 4 * 10-6 * 20= 80 * 10-6 C (b)When the battery is removed the total capacitance ofthe circuit is C = CL + CR = 4 * 10-6 + 4 *10-6 = 8 * 10-6 F Therefore,the total energy stored in the capacitor networkis Utot= (1/2) * C * E2 or Utot= (1/2) * 8 * 10-6 *(20)2 = 1600 * 10-6 J (c)When a dielectric is inserted between the plates ofCR, fully filling the available space as shown then thecapacitance of the capacitor CR is CR'= k *CR= 1.8 * 4 * 10-6 = 7.2 * 10-6F The equivalent capacitance of the two capacitors CLand CR' is C' = CL + CR' = 4 * 10-6 +7.2 * 10-6 = 11.2 * 10-6 F The potential difference across the two capacitors is Q = C' * E' or E' = (Q/C') Here,Q = QL + QR= 80 * 10-6 +80 * 10-6 = 160 * 10-6 C or E' = (160 * 10-6/11.2 * 10-6) = 14.28V Therefore,the new values of the charges on each capacitoris QL'= CL* E' = 4 *10-6 * 14.28 = 57.12 * 10-6 C and QR'= CR'* E' = 7.2 *10-6 * 14.28 = 102.82 * 10-6C (d)The subsequent voltage across CR isVR'= E' = 14.28 V. (e)The total energy stored in the circuit in its finalconfiguration is Utot'= (1/2) * C' * E'2 or Utot'= (1/2) * 11.2 * 10-6 *(14.28)2 or Utot'= 1141.94 * 10-6 J (f)The total energy lost in the capacitor network after thedielectric is inserted is U = Utot - Utot'= 1600 *10-6 - 1141.94 * 10-6 = 458.06 *10-6 J The energy is lost in the capacitor network when thedielectric is inserted.Therefore,the dielectric must be pushed intoCR.This is because when the dielectric is insertedbetween the plates of the capacitor,then there is a change of fluxbetween the capacitorplates.Therefore,the potential difference between theplates of the two capacitors is reduced.Therefore,the energy islost in the capacitor network when the dielectric isinserted.
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