A 9.00 muF parallel-plate capacitor is connected to a 16.0 V battery. (a) How mu

Question

A 9.00 muF parallel-plate capacitor is connected to a 16.0 V battery. (a) How much charge is stored in the capacitor? (b) How much energy is stored in the capacitor? The capacitor is then disconnected from the battery. (c) What is (lie new value of the capacitance when the plate separation is halved? (d) The capacitor still being disconnected from the battery, how much energy is now stored with the new value of the plate separation? (e) Repeal parts (c) and (d) with the capacitor remaining connected to the 16.0 V battery while the plate separation is halved.

Explanation / Answer

a)

C=9.00*10^-6 F, V= 16.0 V

Q=CV = (9.00*10^-6)*(16) = 1.44*10^-4 C

b)

U=1/2*CV^2 = ½*(9.00*10^-6)*(16)^2 = 1.15*10^-3 J

c)

Cold = 0A/dold

Cnew = 0A/dnew

dnew = dold/2

Cnew = 0A/(dold/2) = 20A/dold = 2*9.00*10^-6 F = 18.00*10^-6 F

d)

Charge is always conserved in the capacitor,

Hence, Qnew= Qold

CnewVnew = ColdVold

Vnew = (Cold/Cnew)*Vold = [(9.00*10^-6)/(18.00*10^-6)]*16 = 8.00V

Unew = ½* CnewVnew2 = ½*(18.00*10^-6)*(8)^2 = 5.76*10^-4 J

e)

Cold = 0A/dold

Cnew = 0A/dnew

dnew = dold/2

Cnew = 0A/(dold/2) = 20A/dold = 2*9.00*10^-6 F = 18.00*10^-6 F

f)

Vnew =16.00 V

Unew = ½* CnewVnew2 = ½*(18.00*10^-6)*(16)^2 = 2.3*10^-3 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.