Calculate the capacitance of the capacitor.

Question

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Answer 1

Question

Chapter 6, Problem 40P

(a)

To determine

Calculate the capacitance of the capacitor.

(a)

Expert Solution

Answer to Problem 40P

The capacitance of the capacitor is 59 pF_.

Explanation of Solution

Calculation:

Write the expression for the capacitance.

C=εoSd

Substitute 200 cm2 for S, 3 mm for d, and 10−936π F⋅m−1 for εo.

C=(10−936π F⋅m−1)(200 cm2)3 mm=(10−936π F⋅m−1)(200 ×10−4 m2)3×10−3 m=59×10−12 F=59 pF

Conclusion:

Thus, the capacitance of the capacitor is 59 pF_.

(b)

To determine

Calculate the voltage between the plates.

(b)

Expert Solution

Answer to Problem 40P

The voltage between the plates is 339.3 V_.

Explanation of Solution

Calculation:

Consider the expression for the charge density.

ρs=εoVod

Re-arrange the equation.

Vo=ρsdεo

Substitute 1 μC/m2 for ρs, 3 mm for d, and 10−936π F⋅m−1 for εo.

Vo=(1 μC/m2)(3 mm)10−936π F⋅m−1=(1×10−6 C/m2)(3×10−3 m)10−936π F⋅m−1=(1×10−6 F⋅V/m2)(3×10−3 m)10−936π F⋅m−1 {∵C=F⋅V}=339.3 V

Conclusion:

Thus, the voltage between the plates is 339.3 V_.

(c)

To determine

Calculate the force with which plates attract each other.

(c)

Expert Solution

Answer to Problem 40P

The required force is 1.131 mN_.

Explanation of Solution

Calculation:

Write the expression for the force.

F=ρs2S2εo

Substitute 200 cm2 for S, 1 μC/m2 for ρs, and 10−936π F⋅m−1 for εo.

F=(1 μC/m2)2(200 cm)22(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π C2⋅N−1⋅m−2) {∵F⋅m−1=C2⋅N−1⋅m−2}=1.131 mN

Conclusion:

Thus, the required force is 1.131 mN_.

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Answer 2

Question

Chapter 6, Problem 40P

(a)

To determine

Calculate the capacitance of the capacitor.

(a)

Expert Solution

Answer to Problem 40P

The capacitance of the capacitor is 59 pF_.

Explanation of Solution

Calculation:

Write the expression for the capacitance.

C=εoSd

Substitute 200 cm2 for S, 3 mm for d, and 10−936π F⋅m−1 for εo.

C=(10−936π F⋅m−1)(200 cm2)3 mm=(10−936π F⋅m−1)(200 ×10−4 m2)3×10−3 m=59×10−12 F=59 pF

Conclusion:

Thus, the capacitance of the capacitor is 59 pF_.

(b)

To determine

Calculate the voltage between the plates.

(b)

Expert Solution

Answer to Problem 40P

The voltage between the plates is 339.3 V_.

Explanation of Solution

Calculation:

Consider the expression for the charge density.

ρs=εoVod

Re-arrange the equation.

Vo=ρsdεo

Substitute 1 μC/m2 for ρs, 3 mm for d, and 10−936π F⋅m−1 for εo.

Vo=(1 μC/m2)(3 mm)10−936π F⋅m−1=(1×10−6 C/m2)(3×10−3 m)10−936π F⋅m−1=(1×10−6 F⋅V/m2)(3×10−3 m)10−936π F⋅m−1 {∵C=F⋅V}=339.3 V

Conclusion:

Thus, the voltage between the plates is 339.3 V_.

(c)

To determine

Calculate the force with which plates attract each other.

(c)

Expert Solution

Answer to Problem 40P

The required force is 1.131 mN_.

Explanation of Solution

Calculation:

Write the expression for the force.

F=ρs2S2εo

Substitute 200 cm2 for S, 1 μC/m2 for ρs, and 10−936π F⋅m−1 for εo.

F=(1 μC/m2)2(200 cm)22(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π C2⋅N−1⋅m−2) {∵F⋅m−1=C2⋅N−1⋅m−2}=1.131 mN

Conclusion:

Thus, the required force is 1.131 mN_.

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Answer 3

Question

Chapter 6, Problem 40P

(a)

To determine

Calculate the capacitance of the capacitor.

(a)

Expert Solution

Answer to Problem 40P

The capacitance of the capacitor is 59 pF_.

Explanation of Solution

Calculation:

Write the expression for the capacitance.

C=εoSd

Substitute 200 cm2 for S, 3 mm for d, and 10−936π F⋅m−1 for εo.

C=(10−936π F⋅m−1)(200 cm2)3 mm=(10−936π F⋅m−1)(200 ×10−4 m2)3×10−3 m=59×10−12 F=59 pF

Conclusion:

Thus, the capacitance of the capacitor is 59 pF_.

(b)

To determine

Calculate the voltage between the plates.

(b)

Expert Solution

Answer to Problem 40P

The voltage between the plates is 339.3 V_.

Explanation of Solution

Calculation:

Consider the expression for the charge density.

ρs=εoVod

Re-arrange the equation.

Vo=ρsdεo

Substitute 1 μC/m2 for ρs, 3 mm for d, and 10−936π F⋅m−1 for εo.

Vo=(1 μC/m2)(3 mm)10−936π F⋅m−1=(1×10−6 C/m2)(3×10−3 m)10−936π F⋅m−1=(1×10−6 F⋅V/m2)(3×10−3 m)10−936π F⋅m−1 {∵C=F⋅V}=339.3 V

Conclusion:

Thus, the voltage between the plates is 339.3 V_.

(c)

To determine

Calculate the force with which plates attract each other.

(c)

Expert Solution

Answer to Problem 40P

The required force is 1.131 mN_.

Explanation of Solution

Calculation:

Write the expression for the force.

F=ρs2S2εo

Substitute 200 cm2 for S, 1 μC/m2 for ρs, and 10−936π F⋅m−1 for εo.

F=(1 μC/m2)2(200 cm)22(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π C2⋅N−1⋅m−2) {∵F⋅m−1=C2⋅N−1⋅m−2}=1.131 mN

Conclusion:

Thus, the required force is 1.131 mN_.

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Answer 4

Question

Chapter 6, Problem 40P

(a)

To determine

Calculate the capacitance of the capacitor.

(a)

Expert Solution

Answer to Problem 40P

The capacitance of the capacitor is 59 pF_.

Explanation of Solution

Calculation:

Write the expression for the capacitance.

C=εoSd

Substitute 200 cm2 for S, 3 mm for d, and 10−936π F⋅m−1 for εo.

C=(10−936π F⋅m−1)(200 cm2)3 mm=(10−936π F⋅m−1)(200 ×10−4 m2)3×10−3 m=59×10−12 F=59 pF

Conclusion:

Thus, the capacitance of the capacitor is 59 pF_.

(b)

To determine

Calculate the voltage between the plates.

(b)

Expert Solution

Answer to Problem 40P

The voltage between the plates is 339.3 V_.

Explanation of Solution

Calculation:

Consider the expression for the charge density.

ρs=εoVod

Re-arrange the equation.

Vo=ρsdεo

Substitute 1 μC/m2 for ρs, 3 mm for d, and 10−936π F⋅m−1 for εo.

Vo=(1 μC/m2)(3 mm)10−936π F⋅m−1=(1×10−6 C/m2)(3×10−3 m)10−936π F⋅m−1=(1×10−6 F⋅V/m2)(3×10−3 m)10−936π F⋅m−1 {∵C=F⋅V}=339.3 V

Conclusion:

Thus, the voltage between the plates is 339.3 V_.

(c)

To determine

Calculate the force with which plates attract each other.

(c)

Expert Solution

Answer to Problem 40P

The required force is 1.131 mN_.

Explanation of Solution

Calculation:

Write the expression for the force.

F=ρs2S2εo

Substitute 200 cm2 for S, 1 μC/m2 for ρs, and 10−936π F⋅m−1 for εo.

F=(1 μC/m2)2(200 cm)22(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π C2⋅N−1⋅m−2) {∵F⋅m−1=C2⋅N−1⋅m−2}=1.131 mN

Conclusion:

Thus, the required force is 1.131 mN_.

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Answer 5

Chapter 6, Problem 40P

(a)

To determine

Calculate the capacitance of the capacitor.

(a)

Expert Solution

Answer to Problem 40P

The capacitance of the capacitor is 59 pF_.

Explanation of Solution

Calculation:

Write the expression for the capacitance.

C=εoSd

Substitute 200 cm2 for S, 3 mm for d, and 10−936π F⋅m−1 for εo.

C=(10−936π F⋅m−1)(200 cm2)3 mm=(10−936π F⋅m−1)(200 ×10−4 m2)3×10−3 m=59×10−12 F=59 pF

Conclusion:

Thus, the capacitance of the capacitor is 59 pF_.

(b)

To determine

Calculate the voltage between the plates.

(b)

Expert Solution

Answer to Problem 40P

The voltage between the plates is 339.3 V_.

Explanation of Solution

Calculation:

Consider the expression for the charge density.

ρs=εoVod

Re-arrange the equation.

Vo=ρsdεo

Substitute 1 μC/m2 for ρs, 3 mm for d, and 10−936π F⋅m−1 for εo.

Vo=(1 μC/m2)(3 mm)10−936π F⋅m−1=(1×10−6 C/m2)(3×10−3 m)10−936π F⋅m−1=(1×10−6 F⋅V/m2)(3×10−3 m)10−936π F⋅m−1 {∵C=F⋅V}=339.3 V

Conclusion:

Thus, the voltage between the plates is 339.3 V_.

(c)

To determine

Calculate the force with which plates attract each other.

(c)

Expert Solution

Answer to Problem 40P

The required force is 1.131 mN_.

Explanation of Solution

Calculation:

Write the expression for the force.

F=ρs2S2εo

Substitute 200 cm2 for S, 1 μC/m2 for ρs, and 10−936π F⋅m−1 for εo.

F=(1 μC/m2)2(200 cm)22(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π C2⋅N−1⋅m−2) {∵F⋅m−1=C2⋅N−1⋅m−2}=1.131 mN

Conclusion:

Thus, the required force is 1.131 mN_.

Answer 6

Chapter 6, Problem 40P

(a)

To determine

Calculate the capacitance of the capacitor.

(a)

Expert Solution

Answer to Problem 40P

The capacitance of the capacitor is 59 pF_.

Explanation of Solution

Calculation:

Write the expression for the capacitance.

C=εoSd

Substitute 200 cm2 for S, 3 mm for d, and 10−936π F⋅m−1 for εo.

C=(10−936π F⋅m−1)(200 cm2)3 mm=(10−936π F⋅m−1)(200 ×10−4 m2)3×10−3 m=59×10−12 F=59 pF

Conclusion:

Thus, the capacitance of the capacitor is 59 pF_.

Answer 7

Chapter 6, Problem 40P

(a)

To determine

Calculate the capacitance of the capacitor.

Answer 8

(a)

Expert Solution

Answer to Problem 40P

The capacitance of the capacitor is 59 pF_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

Write the expression for the capacitance.

C=εoSd

Substitute 200 cm2 for S, 3 mm for d, and 10−936π F⋅m−1 for εo.

C=(10−936π F⋅m−1)(200 cm2)3 mm=(10−936π F⋅m−1)(200 ×10−4 m2)3×10−3 m=59×10−12 F=59 pF

Conclusion:

Thus, the capacitance of the capacitor is 59 pF_.

Answer 13

(b)

To determine

Calculate the voltage between the plates.

(b)

Expert Solution

Answer to Problem 40P

The voltage between the plates is 339.3 V_.

Explanation of Solution

Calculation:

Consider the expression for the charge density.

ρs=εoVod

Re-arrange the equation.

Vo=ρsdεo

Substitute 1 μC/m2 for ρs, 3 mm for d, and 10−936π F⋅m−1 for εo.

Vo=(1 μC/m2)(3 mm)10−936π F⋅m−1=(1×10−6 C/m2)(3×10−3 m)10−936π F⋅m−1=(1×10−6 F⋅V/m2)(3×10−3 m)10−936π F⋅m−1 {∵C=F⋅V}=339.3 V

Conclusion:

Thus, the voltage between the plates is 339.3 V_.

Answer 14

(b)

To determine

Calculate the voltage between the plates.

Answer 15

(b)

Expert Solution

Answer to Problem 40P

The voltage between the plates is 339.3 V_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Consider the expression for the charge density.

ρs=εoVod

Re-arrange the equation.

Vo=ρsdεo

Substitute 1 μC/m2 for ρs, 3 mm for d, and 10−936π F⋅m−1 for εo.

Vo=(1 μC/m2)(3 mm)10−936π F⋅m−1=(1×10−6 C/m2)(3×10−3 m)10−936π F⋅m−1=(1×10−6 F⋅V/m2)(3×10−3 m)10−936π F⋅m−1 {∵C=F⋅V}=339.3 V

Conclusion:

Thus, the voltage between the plates is 339.3 V_.

Answer 20

(c)

To determine

Calculate the force with which plates attract each other.

(c)

Expert Solution

Answer to Problem 40P

The required force is 1.131 mN_.

Explanation of Solution

Calculation:

Write the expression for the force.

F=ρs2S2εo

Substitute 200 cm2 for S, 1 μC/m2 for ρs, and 10−936π F⋅m−1 for εo.

F=(1 μC/m2)2(200 cm)22(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π C2⋅N−1⋅m−2) {∵F⋅m−1=C2⋅N−1⋅m−2}=1.131 mN

Conclusion:

Thus, the required force is 1.131 mN_.

Answer 21

(c)

To determine

Calculate the force with which plates attract each other.

Answer 22

(c)

Expert Solution

Answer to Problem 40P

The required force is 1.131 mN_.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

Write the expression for the force.

F=ρs2S2εo

Substitute 200 cm2 for S, 1 μC/m2 for ρs, and 10−936π F⋅m−1 for εo.

F=(1 μC/m2)2(200 cm)22(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π F⋅m−1)=10−12 C2/m4(200×10−4) m2 2(10−936π C2⋅N−1⋅m−2) {∵F⋅m−1=C2⋅N−1⋅m−2}=1.131 mN