Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 18, Problem 136P

(a)

To determine

Find the charge on the upper plate after the capacitor fully charged.

(a)

Expert Solution
Check Mark

Answer to Problem 136P

The charge on the upper plate after the capacitor fully charged is 9.9 nC.

Explanation of Solution

Write the equation for charge on upper plate.

Q=CV=ε0AdV=ε0L2dV (I)

Here, Q is the charge on upper plate, V is potential difference, ε0 permittivity of free space, L is length of the plate and d is the distance between the plates.

Conclusion:

Substitute 10.0 V for V, 8.854×1012 C2/N.m2 for ε0, 0.10 m for L and 89×106 m for d in equation I.

Q=(8.854×1012 C2/N.m2)(0.10 m)2(10.0 V)89×106 m=9.9 nC

Therefore, the charge on the upper plate after the capacitor fully charged is 9.9 nC.

(b)

To determine

Draw the graph between current through the resistor with time.

(b)

Expert Solution
Check Mark

Answer to Problem 136P

The graph between current through the resistor with time is given below.

Explanation of Solution

Write the equation for time constant.

τ=RC (II)

Here, τ is the time constant and R is the resistance.

Substitute the value of C from equation I in equation II.

τ=Rε0L2d (III)

Write the equation for initial current.

I0=VR (IV)

Write the equation for variation of current with time.

I=I0et/τ (V)

Conclusion:

Substitute 0.100×106 Ω for R, 8.854×1012 C2/N.m2 for ε0, 0.10 m for L and 89×106 m for d in equation III.

τ=(0.100×106 Ω)(8.854×1012 C2/N.m2)(0.10 m)289×106 m=9.9×105 s

Substitute 10.0 V for V and 0.100×106 Ω for R in equation IV.

I0=10.0 V0.100×106 Ω=100 μA

Substitute 100 μA for I0 and 9.9×105 s for τ in equation V.

I=(100 μA)et/(9.9×105 s)

Therefore, the graph between current through the resistor with time using the above equation is given below.

Physics, Chapter 18, Problem 136P

(b)

To determine

Find the energy dissipated in whole discharging process.

(b)

Expert Solution
Check Mark

Answer to Problem 136P

The energy dissipated in whole discharging process is 50 nJ.

Explanation of Solution

Write the equation for total energy.

U=12CV2=12(ε0L2d)V2 (VI)

Here, U is the total energy.

Conclusion:

Substitute 10.0 V for V, 8.854×1012 C2/N.m2 for ε0, 0.10 m for L and 89×106 m for d in equation III.

U=12(8.854×1012 C2/N.m2)(0.10 m)2(10.0 V)289×106 m=50 nJ

Therefore, the energy dissipated in whole discharging process is 50 nJ.

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Chapter 18 Solutions

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