Capasitance

Chapter 26. Capacitance

Capacitance
26-1. What is the maximum charge that can be placed on a metal sphere 30 mm in diameter and surrounded by air?
[pic]; Q = 75.0 nC

26-2. How much charge can be placed on a metal sphere of radius 40 mm if it is immersed in transformer oil whose dielectric strength is 16 MV/m?
[pic]; Q = 2.48 (C

26-3. What would be the radius of a metalsphere in air such that it could theoretically hold a charge of one coulomb?
[pic]; r = 54.8 m

26-4. A 28-(F parallel-plate capacitor is connected to a 120-V source of potential difference. How much charge will be stored on this capacitor?
Q = CV = (28 (F)(120 V); Q = 3.36 mC

26-5. A potential difference of 110 V is applied across the plates of a parallel-plate capacitor. If thetotal charge on each plate is 1200 (C, what is the capacitance?
[pic]; C = 10.9 (F

26-6. Find the capacitance of a parallel-plate capacitor is 1600 (C of charge is on each plate when the potential difference is 80 V.
[pic]; C = 20.0 (F

26-7. What potential difference is required to store a charge of 800 (C on a 40-(F capacitor?
[pic]; V = 20.0 V

26-8. Write anequation for the potential at the surface of a sphere of radius r in terms of the permittivity of the surrounding medium. Show that the capacitance of such a sphere is given by C = 4((r.
[pic] [pic]; C = 4(εοr

*26-9. A spherical capacitor has a radius of 50 mm and is surrounded by a medium whose permittivity is 3 x 10-11 C2/N m2. How much charge can be transferred to this sphere by apotential difference of 400 V?
We must replace εο with ε for permittivity of surrounding medium, then
[pic] [pic]; C = 4(ε r
[pic]
[pic]; Q = 4.71 x 10-14 C


Calculating Capacitance
26-10. A 5-(F capacitor has a plate separation of 0.3 mm of air. What will be the charge on each plate for a potential difference of 400 V? What is the area of each plate?
[pic] Q =2000 (C

26-11. The plates of a certain capacitor are 3 mm apart and have an area of 0.04 m2. What is the capacitance if air is the dielectric?
[pic]; C = 118 pF

26-12. A capacitor has plates of area 0.034 m2 and a separation of 2 mm in air. The potential difference between the plates is 200 V. What is the capacitance, and what is the electric field intensity between the plates?How much charge is on each plate?
[pic]; C = 150 pF
[pic]; E = 2.00 x 105 N/C
Q = CV = (150 pF)(400 V); Q = 30.1 (C

26-13. A capacitor of plate area 0.06 m2 and plate separation 4 mm has a potential difference of 300 V when air is the dielectric. What is the capacitance for dielectrics of air (K = 1) and mica (K = 5)?
[pic]; C = 133 pF
[pic]; C = 664 pF
26-14.What is the electric field intensity for mica and for air in Problem 26-13?
[pic][pic]; E0 = 7.50 x 104 V/m
[pic]; E = 1.50 x 104 V/m

26-15. Find the capacitance of a parallel-plate capacitor if the area of each plate is 0.08 m2, the separation of the plates is 4 mm, and the dielectric is (a) air, (b) paraffined paper (K = 2)?
[pic]; C = 177 pF
[pic]; C = 354 pF26-16. Two parallel plates of a capacitor are 4.0 mm apart and the plate area is 0.03 m2. Glass (K = 7.5) is the dielectric, and the plate voltage is 800 V. What is the charge on each plate, and what is the electric field intensity between the plates?
[pic]; C = 498 pF
Q = CV = (498 x 10-12 F)(800 V); Q = 398 nC

*26-17. A parallel-plate capacitor with a capacitance of 2.0 nF isto be constructed with mica (K = 5) as the dielectric, and it must be able to withstand a maximum potential difference of 3000 V. The dielectric strength of mica is 200 MV/m. What is the minimum area the plates of the capacitor can have?
[pic]; [pic]

*16-17. (Cont.) d = 1.50 x 10-5 m; C = 2 nF; K = 5
[pic]; A = 6.78 x 10-4 m2

Capacitors...