ch31-p021

# ch31-p021 - 21. (a) The charge (as a function of time) is...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 21. (a) The charge (as a function of time) is given by q = Q sin ωt , where Q is the maximum charge on the capacitor and ω is the angular frequency of oscillation. A sine function was chosen so that q = 0 at time t = 0. The current (as a function of time) is dq = ωQ cos ωt , dt i= and at t = 0, it is I = ωQ. Since ω = 1/ LC , b Q = I LC = 2.00 A . g c3.00 × 10 Hhc2.70 × 10 Fh = 180 × 10 −3 −6 −4 C. (b) The energy stored in the capacitor is given by UE = q 2 Q 2 sin 2 ωt = 2C 2C and its rate of change is dU E Q 2ω sin ωt cos ωt = dt C bg 1 We use the trigonometric identity cosωt sin ωt = 2 sin 2ωt to write this as dU E ωQ 2 sin 2ωt . = dt 2C bg The greatest rate of change occurs when sin(2ωt) = 1 or 2ωt = π/2 rad. This means t= ππ π = LC = 4ω 4 4 ( 3.00 ×10 −3 H )( 2.70 ×10−6 F ) = 7.07 ×10−5 s. (c) Substituting ω = 2π/T and sin(2ωt) = 1 into dUE/dt = (ωQ2/2C) sin(2ωt), we obtain FG dU IJ H dt K E max 2 πQ 2 πQ 2 = = . TC 2TC c3.00 × 10 Hhc2.70 × 10 Fh = 5.655 × 10 −3 Now T = 2 π LC = 2 π FG dU IJ H dt K = E max −6 c π 180 × 10−4 C . h −4 s, so 2 c5.655 × 10 shc2.70 × 10 Fh = 66.7 W. −4 −6 We note that this is a positive result, indicating that the energy in the capacitor is indeed increasing at t = T/8. ...
View Full Document

## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online