2) A 5uF capacitor is connected to a 6 V batter and allowed to fully charge. It

Question

2) A 5uF capacitor is connected to a 6 V batter and allowed to fully charge. It is then connected to a second uncharged capacitor as shown. A voltmeter is then connected across the pair of capacitors and the reading is 143 Volts. a) Determine the capacitance of the second capacitor and the charge on each capacitor after they are connected. b) The second capacitor has identical geometry to the first (e.g. same area, gap size, for example) but the space between the plates is filled with a dielectric material whereas the first capacitor's gap is filled with air. Which dielectric was used? (See Table 26.1 in Serway or look online.) c) How much charge (total) will flow from the battery if the combination is connected to the 6 V battery starting in its 1.43 V state.

Explanation / Answer

let,

C1=5 uF

V1=6V

===> q1=C1*V1

charge on C1 is q1=(5*10^-6)*6

q1=30uC

a)

let,

uncharged capacitor capacitance is C2

V2=1.43 v

when C1 is connected to C2 ,

charge on capacitor C1 is ,q1'=C1*V1

q1'=5*10^-6*1.43

q1'=7.15 uC

charge on the second capacitor is,

===> q2=q1'-q1

q2=(30-7.15)

q2=22.85 uC

===>

C2=q2/v2

C2=22.85*10^-6/1.43

C2=15.98uF

and

q1'=7.15 uC

q2=22.85 uC

b)

let,

C2=K*eO*A2/d2

C1=eo*A1/d1

and A2=A1 and d2=d1

==>

C2=K*C1

15.98*10^-6=K*5*10^-6

===> K=3.2

dielectric constant, K=3.2( thermo plastic Polyster)

c)

if V=6v

C12=C1+C2

C12=15.98+5

C12=20.98 uF

and

now ,q12=C12*V

q12=20.98*10^-6*6

q12=125.88 uC

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