711_PartUniversity Physics Solution

711_PartUniversity Physics Solution - 24 CAPACITANCE AND...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
24-1 C APACITANCE AND D IELECTRICS 24.1. IDENTIFY: ab Q C V = SET UP: 6 1 F 10 F μ = EXECUTE: 64 (7.28 10 F)(25.0 V) 1.82 10 C 182 C ab QC V −− ==× = × = EVALUATE: One plate has charge Q + and the other has charge Q . 24.2. IDENTIFY and SET UP: 0 A C d = P , Q C V = and VE d = . (a) 2 00 0.00122 m 3.29 pF 0.00328 m A C d == = PP (b) 8 12 4.35 10 C 13.2 kV 3.29 10 F Q V C × = × (c) 3 6 13.2 10 V 4.02 10 V/m 0.00328 m V E d × = × EVALUATE: The electric field is uniform between the plates, at points that aren't close to the edges. 24.3. IDENTIFY and SET UP: It is a parallel-plate air capacitor, so we can apply the equations of Sections 24.1. EXECUTE: (a) 6 12 0.148 10 C so 604 V 245 10 F ab ab QQ CV VC × = = × (b) 0 so A C d = P () ( ) 12 3 32 2 12 2 2 0 245 10 F 0.328 10 m 9.08 10 m 90.8 cm 8.854 10 C / N m Cd A ×× = × = ×⋅ P (c) 6 3 604 V so 1.84 10 V/m 0.328 10 m ab ab V dE d = = × × (d) 0 so E σ = P ( ) 61 2 2 2 5 2 0 1.84 10 V/m 8.854 10 C / N m 1.63 10 C/m E × × ⋅=× P EVALUATE: We could also calculate directly as Q/A . 6 52 1.63 10 C/m , Q A × × which checks. 24.4. IDENTIFY: 0 A C d = P when there is air between the plates. SET UP: 22 (3.0 10 m) A is the area of each plate. EXECUTE: 12 2 2 12 3 (8.854 10 F/m)(3.0 10 m) 1.59 10 F 1.59 pF 5.0 10 m C × = × EVALUATE: C increases when A increases and C increases when d decreases. 24.5. IDENTIFY: ab Q C V = . 0 A C d = P . SET UP: When the capacitor is connected to the battery, 12.0 V ab V = . EXECUTE: (a) (10.0 10 F)(12.0 V) 1.20 10 C 120 C ab V = × = (b) When d is doubled C is halved, so Q is halved. 60 C Q = . (c) If r is doubled, A increases by a factor of 4. C increases by a factor of 4 and Q increases by a factor of 4. 480 C. Q = EVALUATE: When the plates are moved apart, less charge on the plates is required to produce the same potential difference. With the separation of the plates constant, the electric field must remain constant to produce the same potential difference. The electric field depends on the surface charge density, . To produce the same , more charge is required when the area increases. 24
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
24-2 Chapter 24 24.6. IDENTIFY: ab Q C V = . 0 A C d = P . SET UP: When the capacitor is connected to the battery, enough charge flows onto the plates to make 12.0 V. ab V = EXECUTE: (a) 12.0 V (b) (i) When d is doubled, C is halved. ab Q V C = and Q is constant, so V doubles. 24.0 V V = . (ii) When r is doubled, A increases by a factor of 4. V decreases by a factor of 4 and 3.0 V V = .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

711_PartUniversity Physics Solution - 24 CAPACITANCE AND...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online