Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 25, Problem 99P

(a)

To determine

The charge on the capacitor plate.

(a)

Expert Solution
Check Mark

Answer to Problem 99P

The charge on the capacitor plate is 5.69μC .

Explanation of Solution

Given:

The value of emf is ε=6.00V .

The value of capacitance is C=1.50μF .

The resistance is R=2.00MΩ .

Formula used:

The expression for the charge as a function of time is given as,

  Q(t)=Cε(1et/τ)

Calculation:

The charge on the capacitor plate after t=τsec is calculated as,

  Q(t)=(1.50μF)(6.00V)(1e τ/τ )=(1.50μF)(6.00V)(1e 1)=5.689μC5.69μC

Conclusion:

Therefore, the charge on the capacitor plate is 5.69μC .

(b)

To determine

The rate of increasing charge.

(b)

Expert Solution
Check Mark

Answer to Problem 99P

The rate of increasing charge is 1.10μC/s .

Explanation of Solution

Formula used:

The expression for rate of increasing charge is,

  dQdt=εRet/τ

Calculation:

The rate of increasing charge after t=τsec is calculated as,

  dQdt=( 6.00V 2.00MΩ)eτ/τ=( 6.00V 2.00× 10 6 Ω)e1=1.104×106C/s=1.10μC/s ..... (1)

Conclusion:

Therefore, the rate of increasing charge is 1.10μC/s .

(c)

To determine

The current after t=τsec .

(c)

Expert Solution
Check Mark

Answer to Problem 99P

The current after t=τsec is 1.10μA .

Explanation of Solution

Formula used:

The expression for current is,

  I(t)=dQdt ..... (2)

Calculation:

From equation (1) and (2), the current after t=τsec is calculated as,

  I(t)=1.10μC/s=1.10μA

Conclusion:

Therefore, the current after t=τsec is 1.10μA .

(d)

To determine

The power supplied by battery.

(d)

Expert Solution
Check Mark

Answer to Problem 99P

The power supplied by battery is 6.62μW .

Explanation of Solution

Formula used:

The expression for power is,

  P(t)=εI(t)

Calculation:

The power supplied by battery after t=τsec is calculated as,

  P(t)=(6.00V)(1.104μA)=6.62μW

Conclusion:

Therefore, the power supplied by battery is 6.62μW .

(e)

To determine

The power delivered to resistor.

(e)

Expert Solution
Check Mark

Answer to Problem 99P

The power delivered to resistor is 2.44μW .

Explanation of Solution

Formula used:

The expression for power delivered to resistor is,

  PR(t)=I2(t)R

Calculation:

The power supplied by battery after t=τsec is calculated as,

  PR(t)=(1.104μA)2(2.00MΩ)=(1.104× 10 6A)2(2.00× 106Ω)=2.437μW=2.44μW

Conclusion:

Therefore, the power delivered to resistor is 2.44μW .

(f)

To determine

The rate of stored energy.

(f)

Expert Solution
Check Mark

Answer to Problem 99P

The rate of stored energy is 4.19μW .

Explanation of Solution

Formula used:

The expression for stored energy is,

  U(t)=12CQ2(t) ..... (3)

Calculation:

Differentiate equation (3) with respect to time,

  dU(t)dt=12C{2Q(t)}dQ(t)dt=Q(t)CI(t)

The rate of stored energy is calculated as,

  dU(t)dt=5.69μC1.50μF(1.104μA)=4.187μW=4.19μW

Conclusion:

Therefore, the rate of stored energy is 4.19μW .

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Chapter 25 Solutions

Physics for Scientists and Engineers

Ch. 25 - Prob. 11PCh. 25 - Prob. 12PCh. 25 - Prob. 13PCh. 25 - Prob. 14PCh. 25 - Prob. 15PCh. 25 - Prob. 16PCh. 25 - Prob. 17PCh. 25 - Prob. 18PCh. 25 - Prob. 19PCh. 25 - Prob. 20PCh. 25 - Prob. 21PCh. 25 - Prob. 22PCh. 25 - Prob. 23PCh. 25 - Prob. 24PCh. 25 - Prob. 25PCh. 25 - Prob. 26PCh. 25 - Prob. 27PCh. 25 - Prob. 28PCh. 25 - Prob. 29PCh. 25 - Prob. 30PCh. 25 - Prob. 31PCh. 25 - Prob. 32PCh. 25 - Prob. 33PCh. 25 - Prob. 34PCh. 25 - Prob. 35PCh. 25 - Prob. 36PCh. 25 - Prob. 37PCh. 25 - Prob. 38PCh. 25 - Prob. 39PCh. 25 - Prob. 40PCh. 25 - Prob. 41PCh. 25 - Prob. 42PCh. 25 - Prob. 43PCh. 25 - Prob. 44PCh. 25 - Prob. 45PCh. 25 - Prob. 46PCh. 25 - Prob. 47PCh. 25 - Prob. 48PCh. 25 - Prob. 49PCh. 25 - Prob. 50PCh. 25 - Prob. 51PCh. 25 - Prob. 52PCh. 25 - Prob. 53PCh. 25 - Prob. 54PCh. 25 - Prob. 55PCh. 25 - Prob. 56PCh. 25 - Prob. 57PCh. 25 - Prob. 58PCh. 25 - Prob. 59PCh. 25 - Prob. 60PCh. 25 - Prob. 61PCh. 25 - Prob. 62PCh. 25 - Prob. 63PCh. 25 - Prob. 64PCh. 25 - Prob. 65PCh. 25 - Prob. 66PCh. 25 - Prob. 67PCh. 25 - Prob. 68PCh. 25 - Prob. 69PCh. 25 - Prob. 70PCh. 25 - Prob. 71PCh. 25 - Prob. 72PCh. 25 - Prob. 73PCh. 25 - Prob. 74PCh. 25 - Prob. 75PCh. 25 - Prob. 76PCh. 25 - Prob. 77PCh. 25 - Prob. 78PCh. 25 - Prob. 79PCh. 25 - Prob. 80PCh. 25 - Prob. 81PCh. 25 - Prob. 82PCh. 25 - Prob. 83PCh. 25 - Prob. 84PCh. 25 - Prob. 85PCh. 25 - Prob. 86PCh. 25 - Prob. 87PCh. 25 - Prob. 88PCh. 25 - Prob. 89PCh. 25 - Prob. 90PCh. 25 - Prob. 91PCh. 25 - Prob. 92PCh. 25 - Prob. 93PCh. 25 - Prob. 94PCh. 25 - Prob. 95PCh. 25 - Prob. 96PCh. 25 - Prob. 97PCh. 25 - Prob. 98PCh. 25 - Prob. 99PCh. 25 - Prob. 100PCh. 25 - Prob. 101PCh. 25 - Prob. 102PCh. 25 - Prob. 103PCh. 25 - Prob. 104PCh. 25 - Prob. 105PCh. 25 - Prob. 106PCh. 25 - Prob. 107PCh. 25 - Prob. 108PCh. 25 - Prob. 109PCh. 25 - Prob. 110PCh. 25 - Prob. 111PCh. 25 - Prob. 112PCh. 25 - Prob. 113PCh. 25 - Prob. 114PCh. 25 - Prob. 115PCh. 25 - Prob. 116PCh. 25 - Prob. 117PCh. 25 - Prob. 118PCh. 25 - Prob. 119PCh. 25 - Prob. 120P
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