(c) Capacitors Cs 15 AuF and C4-25 AuF are initially charged as a series combina

Question

(c) Capacitors Cs 15 AuF and C4-25 AuF are initially charged as a series combination across a 60-Vbattery. After charging, these capacitors are then disconnected from the voltage source and reconnected with the two positive plates connected and the two negative plates connected. Find the resulting energy stored on each capacitor at steady tate. U5.9 m; U-9.9 m] 3. (a) A parallel-plate capacitor, having plate area of 4.0 m2 and are separated by d-4.0 mm, is initially across a 100-V battery. When the capacitor is fully charged, find the: (i) Capacitance;i) Charge on the capacitor; iii) Electric field strength; and (iv) Electrostatic energy stored in the capacitor. [(i) 8.85 nF, (ii) 0.885 ?C; (iii) 25 kVin; (iv) 44.3 ?J] (b) After this capacitor above has been fully charged, the battery is disconnected. A dielectric stack of paper (? 3.7) is carefully inserted between the capacitor plates so that it completely fills the space between the plates without changing the plate separation of the capacitor. Assume that the charge in the capacitor remains constant. With the dielectric inserted in the capacitor, find the new values of the: (i) Capacitance (ii Electric field strength; (ii) Voltage between the plates; and (iv) Induced charges that appear on the dielectric. (v) Calculate the difference in the electrostatic energy in the capacitor before and after the dielectric was inserted. (vi) What factors account for the energy change? [(i) 32.8 nF: (ii) 676 kVm; (iii) 27 V: (iv) 0.646 C; (v)-32.3 ,?; (vi) ?]

Explanation / Answer

c) C3 = 15 uF

C4 = 25 uF

initially charged in series combination and hence have same charge q

V = 60 V

hence

q = Ceff*V

Ceff = C3*C4/(C3 + C4) = 9.375 uF

hence

q = 562.5uC

hence

E1 = 0.5*Q^2(1/C3 + 1/C4) = 0.016875 J

now, when they are disconnected and connected with +ve plates and -ve plates together

net +ve charge = 2q

hence

for parallel connection the potential drop is sameacross capacitors

hence

q3/C3 = q4/C4

also

q3 + q4 = 2q

q4(1 + C3/C4) = 2q

q4 = 703.125 uC

q3 = 421.875 uC

hence

E2 = 0.5q3^2/C3 + 0.5q4^2/C4 = 0.0158203125 J

Ufinal = 15.8203125 mJ

U3 = 0.5*q3^2/C3 = 5.9326171875 mJ

U4 = 9.8876953125mJ

3 a) A = 4 m^2

d = 4 mm

V = 100 V

i. capacitance = C

C = A*epsilon/d = A*4*pi*epsilon/4*pi*d = A/4*pi*k*d

k = 8.98*10^9

hence

C = 8.8616*10^-9 F = 8.8616 nF

ii. charge, Q = CV = 8.8616*10^-7 C

iii. E = V/d = 25000V/m

iv. U = 0.5CV^2 = 44.308 uJ

b. i. new capacitance = k*C = 32.78792 nF

ii. E = V/d = Q/k*C*d = 6756.756756756756 V/m

iii. V = E*d = 27.027027027027 V

iv. induced charge = Q(1 - 1/k) = 6.46657*10^-7 C

v. Uf = 0.5Q^2/kC = 11.9751351 uJ

U - Uf = 32.33286486 uJ

vi. the induced charges need energy and hence account for energy change

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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