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

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