The radius and the mass of the drop used in the experiment.

Question

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Answer 1

Question

Chapter 4, Problem 6P

(a)

To determine

The radius and the mass of the drop used in the experiment.

(a)

Expert Solution

Answer to Problem 6P

The radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

Explanation of Solution

Write the expression for the radius of the oil drop in the Millikan’s experiment.

a=9ηv2ρg (I)

Here, η is the viscosity of the liquid, v is the velocity of the oil drop, ρ is the density and g is the acceleration due to gravity.

The mass of the oil drop is given by,

m=ρV (II)

Here, V is the volume, m is the mass of the oil drop.

The volume of the oil drop is given by,

V=43πa3 (III)

Use equation (III) in (II),

m=ρ(43)πa3 (IV)

The average speed of the oil drop is given as,

v=ΔyΔt=4.0×10−3m15.9s=2.52×10−4m/s

Conclusion:

Substitute 1.81×10−5kg/ms for η, 2.52×10−4m/s for v, 800kg/m3 for ρ and 9.81m/s2 for g in equation (I) to find a.

a=(9(1.81×10−5kg/ms)(2.52×10−4m/s)2(800kg/m3)(9.81m/s2))1/2=1.62×10−6 m

Substitute 1.33 for ρ, 1.62×10−6 m for a in equation (IV) to find m.

m=(1.33)(800kg/m3)(π)(1.62×10−6m)3=1.42×10−14 kg

Therefore, the radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

(b)

To determine

The charge on each drop and show that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(b)

Expert Solution

Answer to Problem 6P

It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Explanation of Solution

The charge on the droplet is given by,

q1=mgE(v+v1′)v (V)

Here, v is the velocity of fall, v′ is the velocity of rise, E is the electric field, v′ is the upward terminal speed in the presence of the electric field, v is the downward terminal velocity in the absence of the electric field.

The expression for the electric field is given by,

E=Vd (VI)

Conclusion:

Using the times given, the various v′ can be calculated as,

v1′=4×10−3 m36.0 s=1.11×10−4m/sv2′=4×10−3m17.3 s=2.31×10−4m/s

v3′=4×10−3 m24.0 s=1.67×10−4m/sv4′=4×10−3 m11.4 s =3.51×10−4m/s

v5′=4×10−3 m7.54 s=5.31×10−4m/s

The corresponding charges can be written as,

q1=(6.97×10−19C)(2.52+1.11)2.52=10.0×10−19Cq2=13.4×10−19 C

q3=11.6×10−19Cq4=16.7×10−19Cq5=21.6×10−19C

Therefore, It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(c)

To determine

The electronic charge from the data.

(c)

Expert Solution

Answer to Problem 6P

The average value is 1.67×10−19 C_.

Explanation of Solution

To find an integer n1 such that 1.5×10−19<qn1<2.0×10−19

q16=1.67×10−19Cq28=1.68×10−19Cq37=1.66×10−19Cq410=1.67×10−19C

q513=1.66×10−19C

The amount of charge gained or lost is,

q2−q1=3.4×10−19Cq3−q1=1.6×10−19Cq2−q3=1.8×10−19Cq4−q1=6.7×10−19C

q4−q3=5.1×10−19Cq4−q2=3.4×10−19Cq5−q1=11.6×10−19Cq5−q4=4.9×10−19C

q5−q4=4.9×10−19Cq5−q3=10.0×10−19C

The integers that yield a value of e between 1.5 and 2.0×10−19 C.

q2−q12=1.70×10−19Cq3−q11=1.60×10−19Cq2−q31=1.80×10−19Cq4−q14=1.68×10−19C

q4−q33=1.70×10−19Cq4−q22=1.65×10−19Cq5−q17=1.66×10−19Cq5−q43=1.63×10−19C

q5−q36=1.67×10−19C

The average of all values will be 1.67×10−19 C.

Conclusion:

Therefore, The average value is 1.67×10−19 C_.

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Answer 2

Question

Chapter 4, Problem 6P

(a)

To determine

The radius and the mass of the drop used in the experiment.

(a)

Expert Solution

Answer to Problem 6P

The radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

Explanation of Solution

Write the expression for the radius of the oil drop in the Millikan’s experiment.

a=9ηv2ρg (I)

Here, η is the viscosity of the liquid, v is the velocity of the oil drop, ρ is the density and g is the acceleration due to gravity.

The mass of the oil drop is given by,

m=ρV (II)

Here, V is the volume, m is the mass of the oil drop.

The volume of the oil drop is given by,

V=43πa3 (III)

Use equation (III) in (II),

m=ρ(43)πa3 (IV)

The average speed of the oil drop is given as,

v=ΔyΔt=4.0×10−3m15.9s=2.52×10−4m/s

Conclusion:

Substitute 1.81×10−5kg/ms for η, 2.52×10−4m/s for v, 800kg/m3 for ρ and 9.81m/s2 for g in equation (I) to find a.

a=(9(1.81×10−5kg/ms)(2.52×10−4m/s)2(800kg/m3)(9.81m/s2))1/2=1.62×10−6 m

Substitute 1.33 for ρ, 1.62×10−6 m for a in equation (IV) to find m.

m=(1.33)(800kg/m3)(π)(1.62×10−6m)3=1.42×10−14 kg

Therefore, the radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

(b)

To determine

The charge on each drop and show that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(b)

Expert Solution

Answer to Problem 6P

It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Explanation of Solution

The charge on the droplet is given by,

q1=mgE(v+v1′)v (V)

Here, v is the velocity of fall, v′ is the velocity of rise, E is the electric field, v′ is the upward terminal speed in the presence of the electric field, v is the downward terminal velocity in the absence of the electric field.

The expression for the electric field is given by,

E=Vd (VI)

Conclusion:

Using the times given, the various v′ can be calculated as,

v1′=4×10−3 m36.0 s=1.11×10−4m/sv2′=4×10−3m17.3 s=2.31×10−4m/s

v3′=4×10−3 m24.0 s=1.67×10−4m/sv4′=4×10−3 m11.4 s =3.51×10−4m/s

v5′=4×10−3 m7.54 s=5.31×10−4m/s

The corresponding charges can be written as,

q1=(6.97×10−19C)(2.52+1.11)2.52=10.0×10−19Cq2=13.4×10−19 C

q3=11.6×10−19Cq4=16.7×10−19Cq5=21.6×10−19C

Therefore, It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(c)

To determine

The electronic charge from the data.

(c)

Expert Solution

Answer to Problem 6P

The average value is 1.67×10−19 C_.

Explanation of Solution

To find an integer n1 such that 1.5×10−19<qn1<2.0×10−19

q16=1.67×10−19Cq28=1.68×10−19Cq37=1.66×10−19Cq410=1.67×10−19C

q513=1.66×10−19C

The amount of charge gained or lost is,

q2−q1=3.4×10−19Cq3−q1=1.6×10−19Cq2−q3=1.8×10−19Cq4−q1=6.7×10−19C

q4−q3=5.1×10−19Cq4−q2=3.4×10−19Cq5−q1=11.6×10−19Cq5−q4=4.9×10−19C

q5−q4=4.9×10−19Cq5−q3=10.0×10−19C

The integers that yield a value of e between 1.5 and 2.0×10−19 C.

q2−q12=1.70×10−19Cq3−q11=1.60×10−19Cq2−q31=1.80×10−19Cq4−q14=1.68×10−19C

q4−q33=1.70×10−19Cq4−q22=1.65×10−19Cq5−q17=1.66×10−19Cq5−q43=1.63×10−19C

q5−q36=1.67×10−19C

The average of all values will be 1.67×10−19 C.

Conclusion:

Therefore, The average value is 1.67×10−19 C_.

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Answer 3

Question

Chapter 4, Problem 6P

(a)

To determine

The radius and the mass of the drop used in the experiment.

(a)

Expert Solution

Answer to Problem 6P

The radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

Explanation of Solution

Write the expression for the radius of the oil drop in the Millikan’s experiment.

a=9ηv2ρg (I)

Here, η is the viscosity of the liquid, v is the velocity of the oil drop, ρ is the density and g is the acceleration due to gravity.

The mass of the oil drop is given by,

m=ρV (II)

Here, V is the volume, m is the mass of the oil drop.

The volume of the oil drop is given by,

V=43πa3 (III)

Use equation (III) in (II),

m=ρ(43)πa3 (IV)

The average speed of the oil drop is given as,

v=ΔyΔt=4.0×10−3m15.9s=2.52×10−4m/s

Conclusion:

Substitute 1.81×10−5kg/ms for η, 2.52×10−4m/s for v, 800kg/m3 for ρ and 9.81m/s2 for g in equation (I) to find a.

a=(9(1.81×10−5kg/ms)(2.52×10−4m/s)2(800kg/m3)(9.81m/s2))1/2=1.62×10−6 m

Substitute 1.33 for ρ, 1.62×10−6 m for a in equation (IV) to find m.

m=(1.33)(800kg/m3)(π)(1.62×10−6m)3=1.42×10−14 kg

Therefore, the radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

(b)

To determine

The charge on each drop and show that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(b)

Expert Solution

Answer to Problem 6P

It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Explanation of Solution

The charge on the droplet is given by,

q1=mgE(v+v1′)v (V)

Here, v is the velocity of fall, v′ is the velocity of rise, E is the electric field, v′ is the upward terminal speed in the presence of the electric field, v is the downward terminal velocity in the absence of the electric field.

The expression for the electric field is given by,

E=Vd (VI)

Conclusion:

Using the times given, the various v′ can be calculated as,

v1′=4×10−3 m36.0 s=1.11×10−4m/sv2′=4×10−3m17.3 s=2.31×10−4m/s

v3′=4×10−3 m24.0 s=1.67×10−4m/sv4′=4×10−3 m11.4 s =3.51×10−4m/s

v5′=4×10−3 m7.54 s=5.31×10−4m/s

The corresponding charges can be written as,

q1=(6.97×10−19C)(2.52+1.11)2.52=10.0×10−19Cq2=13.4×10−19 C

q3=11.6×10−19Cq4=16.7×10−19Cq5=21.6×10−19C

Therefore, It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(c)

To determine

The electronic charge from the data.

(c)

Expert Solution

Answer to Problem 6P

The average value is 1.67×10−19 C_.

Explanation of Solution

To find an integer n1 such that 1.5×10−19<qn1<2.0×10−19

q16=1.67×10−19Cq28=1.68×10−19Cq37=1.66×10−19Cq410=1.67×10−19C

q513=1.66×10−19C

The amount of charge gained or lost is,

q2−q1=3.4×10−19Cq3−q1=1.6×10−19Cq2−q3=1.8×10−19Cq4−q1=6.7×10−19C

q4−q3=5.1×10−19Cq4−q2=3.4×10−19Cq5−q1=11.6×10−19Cq5−q4=4.9×10−19C

q5−q4=4.9×10−19Cq5−q3=10.0×10−19C

The integers that yield a value of e between 1.5 and 2.0×10−19 C.

q2−q12=1.70×10−19Cq3−q11=1.60×10−19Cq2−q31=1.80×10−19Cq4−q14=1.68×10−19C

q4−q33=1.70×10−19Cq4−q22=1.65×10−19Cq5−q17=1.66×10−19Cq5−q43=1.63×10−19C

q5−q36=1.67×10−19C

The average of all values will be 1.67×10−19 C.

Conclusion:

Therefore, The average value is 1.67×10−19 C_.

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Answer 4

Question

Chapter 4, Problem 6P

(a)

To determine

The radius and the mass of the drop used in the experiment.

(a)

Expert Solution

Answer to Problem 6P

The radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

Explanation of Solution

Write the expression for the radius of the oil drop in the Millikan’s experiment.

a=9ηv2ρg (I)

Here, η is the viscosity of the liquid, v is the velocity of the oil drop, ρ is the density and g is the acceleration due to gravity.

The mass of the oil drop is given by,

m=ρV (II)

Here, V is the volume, m is the mass of the oil drop.

The volume of the oil drop is given by,

V=43πa3 (III)

Use equation (III) in (II),

m=ρ(43)πa3 (IV)

The average speed of the oil drop is given as,

v=ΔyΔt=4.0×10−3m15.9s=2.52×10−4m/s

Conclusion:

Substitute 1.81×10−5kg/ms for η, 2.52×10−4m/s for v, 800kg/m3 for ρ and 9.81m/s2 for g in equation (I) to find a.

a=(9(1.81×10−5kg/ms)(2.52×10−4m/s)2(800kg/m3)(9.81m/s2))1/2=1.62×10−6 m

Substitute 1.33 for ρ, 1.62×10−6 m for a in equation (IV) to find m.

m=(1.33)(800kg/m3)(π)(1.62×10−6m)3=1.42×10−14 kg

Therefore, the radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

(b)

To determine

The charge on each drop and show that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(b)

Expert Solution

Answer to Problem 6P

It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Explanation of Solution

The charge on the droplet is given by,

q1=mgE(v+v1′)v (V)

Here, v is the velocity of fall, v′ is the velocity of rise, E is the electric field, v′ is the upward terminal speed in the presence of the electric field, v is the downward terminal velocity in the absence of the electric field.

The expression for the electric field is given by,

E=Vd (VI)

Conclusion:

Using the times given, the various v′ can be calculated as,

v1′=4×10−3 m36.0 s=1.11×10−4m/sv2′=4×10−3m17.3 s=2.31×10−4m/s

v3′=4×10−3 m24.0 s=1.67×10−4m/sv4′=4×10−3 m11.4 s =3.51×10−4m/s

v5′=4×10−3 m7.54 s=5.31×10−4m/s

The corresponding charges can be written as,

q1=(6.97×10−19C)(2.52+1.11)2.52=10.0×10−19Cq2=13.4×10−19 C

q3=11.6×10−19Cq4=16.7×10−19Cq5=21.6×10−19C

Therefore, It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(c)

To determine

The electronic charge from the data.

(c)

Expert Solution

Answer to Problem 6P

The average value is 1.67×10−19 C_.

Explanation of Solution

To find an integer n1 such that 1.5×10−19<qn1<2.0×10−19

q16=1.67×10−19Cq28=1.68×10−19Cq37=1.66×10−19Cq410=1.67×10−19C

q513=1.66×10−19C

The amount of charge gained or lost is,

q2−q1=3.4×10−19Cq3−q1=1.6×10−19Cq2−q3=1.8×10−19Cq4−q1=6.7×10−19C

q4−q3=5.1×10−19Cq4−q2=3.4×10−19Cq5−q1=11.6×10−19Cq5−q4=4.9×10−19C

q5−q4=4.9×10−19Cq5−q3=10.0×10−19C

The integers that yield a value of e between 1.5 and 2.0×10−19 C.

q2−q12=1.70×10−19Cq3−q11=1.60×10−19Cq2−q31=1.80×10−19Cq4−q14=1.68×10−19C

q4−q33=1.70×10−19Cq4−q22=1.65×10−19Cq5−q17=1.66×10−19Cq5−q43=1.63×10−19C

q5−q36=1.67×10−19C

The average of all values will be 1.67×10−19 C.

Conclusion:

Therefore, The average value is 1.67×10−19 C_.

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Answer 5

Chapter 4, Problem 6P

(a)

To determine

The radius and the mass of the drop used in the experiment.

(a)

Expert Solution

Answer to Problem 6P

The radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

Explanation of Solution

Write the expression for the radius of the oil drop in the Millikan’s experiment.

a=9ηv2ρg (I)

Here, η is the viscosity of the liquid, v is the velocity of the oil drop, ρ is the density and g is the acceleration due to gravity.

The mass of the oil drop is given by,

m=ρV (II)

Here, V is the volume, m is the mass of the oil drop.

The volume of the oil drop is given by,

V=43πa3 (III)

Use equation (III) in (II),

m=ρ(43)πa3 (IV)

The average speed of the oil drop is given as,

v=ΔyΔt=4.0×10−3m15.9s=2.52×10−4m/s

Conclusion:

Substitute 1.81×10−5kg/ms for η, 2.52×10−4m/s for v, 800kg/m3 for ρ and 9.81m/s2 for g in equation (I) to find a.

a=(9(1.81×10−5kg/ms)(2.52×10−4m/s)2(800kg/m3)(9.81m/s2))1/2=1.62×10−6 m

Substitute 1.33 for ρ, 1.62×10−6 m for a in equation (IV) to find m.

m=(1.33)(800kg/m3)(π)(1.62×10−6m)3=1.42×10−14 kg

Therefore, the radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

(b)

To determine

The charge on each drop and show that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(b)

Expert Solution

Answer to Problem 6P

It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Explanation of Solution

The charge on the droplet is given by,

q1=mgE(v+v1′)v (V)

Here, v is the velocity of fall, v′ is the velocity of rise, E is the electric field, v′ is the upward terminal speed in the presence of the electric field, v is the downward terminal velocity in the absence of the electric field.

The expression for the electric field is given by,

E=Vd (VI)

Conclusion:

Using the times given, the various v′ can be calculated as,

v1′=4×10−3 m36.0 s=1.11×10−4m/sv2′=4×10−3m17.3 s=2.31×10−4m/s

v3′=4×10−3 m24.0 s=1.67×10−4m/sv4′=4×10−3 m11.4 s =3.51×10−4m/s

v5′=4×10−3 m7.54 s=5.31×10−4m/s

The corresponding charges can be written as,

q1=(6.97×10−19C)(2.52+1.11)2.52=10.0×10−19Cq2=13.4×10−19 C

q3=11.6×10−19Cq4=16.7×10−19Cq5=21.6×10−19C

Therefore, It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(c)

To determine

The electronic charge from the data.

(c)

Expert Solution

Answer to Problem 6P

The average value is 1.67×10−19 C_.

Explanation of Solution

To find an integer n1 such that 1.5×10−19<qn1<2.0×10−19

q16=1.67×10−19Cq28=1.68×10−19Cq37=1.66×10−19Cq410=1.67×10−19C

q513=1.66×10−19C

The amount of charge gained or lost is,

q2−q1=3.4×10−19Cq3−q1=1.6×10−19Cq2−q3=1.8×10−19Cq4−q1=6.7×10−19C

q4−q3=5.1×10−19Cq4−q2=3.4×10−19Cq5−q1=11.6×10−19Cq5−q4=4.9×10−19C

q5−q4=4.9×10−19Cq5−q3=10.0×10−19C

The integers that yield a value of e between 1.5 and 2.0×10−19 C.

q2−q12=1.70×10−19Cq3−q11=1.60×10−19Cq2−q31=1.80×10−19Cq4−q14=1.68×10−19C

q4−q33=1.70×10−19Cq4−q22=1.65×10−19Cq5−q17=1.66×10−19Cq5−q43=1.63×10−19C

q5−q36=1.67×10−19C

The average of all values will be 1.67×10−19 C.

Conclusion:

Therefore, The average value is 1.67×10−19 C_.

Answer 6

Chapter 4, Problem 6P

(a)

To determine

The radius and the mass of the drop used in the experiment.

(a)

Expert Solution

Answer to Problem 6P

The radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

Explanation of Solution

Write the expression for the radius of the oil drop in the Millikan’s experiment.

a=9ηv2ρg (I)

Here, η is the viscosity of the liquid, v is the velocity of the oil drop, ρ is the density and g is the acceleration due to gravity.

The mass of the oil drop is given by,

m=ρV (II)

Here, V is the volume, m is the mass of the oil drop.

The volume of the oil drop is given by,

V=43πa3 (III)

Use equation (III) in (II),

m=ρ(43)πa3 (IV)

The average speed of the oil drop is given as,

v=ΔyΔt=4.0×10−3m15.9s=2.52×10−4m/s

Conclusion:

Substitute 1.81×10−5kg/ms for η, 2.52×10−4m/s for v, 800kg/m3 for ρ and 9.81m/s2 for g in equation (I) to find a.

a=(9(1.81×10−5kg/ms)(2.52×10−4m/s)2(800kg/m3)(9.81m/s2))1/2=1.62×10−6 m

Substitute 1.33 for ρ, 1.62×10−6 m for a in equation (IV) to find m.

m=(1.33)(800kg/m3)(π)(1.62×10−6m)3=1.42×10−14 kg

Therefore, the radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

Answer 7

Chapter 4, Problem 6P

(a)

To determine

The radius and the mass of the drop used in the experiment.

Answer 8

(a)

Expert Solution

Answer to Problem 6P

The radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the radius of the oil drop in the Millikan’s experiment.

a=9ηv2ρg (I)

Here, η is the viscosity of the liquid, v is the velocity of the oil drop, ρ is the density and g is the acceleration due to gravity.

The mass of the oil drop is given by,

m=ρV (II)

Here, V is the volume, m is the mass of the oil drop.

The volume of the oil drop is given by,

V=43πa3 (III)

Use equation (III) in (II),

m=ρ(43)πa3 (IV)

The average speed of the oil drop is given as,

v=ΔyΔt=4.0×10−3m15.9s=2.52×10−4m/s

Conclusion:

Substitute 1.81×10−5kg/ms for η, 2.52×10−4m/s for v, 800kg/m3 for ρ and 9.81m/s2 for g in equation (I) to find a.

a=(9(1.81×10−5kg/ms)(2.52×10−4m/s)2(800kg/m3)(9.81m/s2))1/2=1.62×10−6 m

Substitute 1.33 for ρ, 1.62×10−6 m for a in equation (IV) to find m.

m=(1.33)(800kg/m3)(π)(1.62×10−6m)3=1.42×10−14 kg

Therefore, the radius is 1.62×10−6 m_ and the mass is 1.42×10−14 kg_.

Answer 13

(b)

To determine

The charge on each drop and show that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

(b)

Expert Solution

Answer to Problem 6P

It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Explanation of Solution

The charge on the droplet is given by,

q1=mgE(v+v1′)v (V)

Here, v is the velocity of fall, v′ is the velocity of rise, E is the electric field, v′ is the upward terminal speed in the presence of the electric field, v is the downward terminal velocity in the absence of the electric field.

The expression for the electric field is given by,

E=Vd (VI)

Conclusion:

Using the times given, the various v′ can be calculated as,

v1′=4×10−3 m36.0 s=1.11×10−4m/sv2′=4×10−3m17.3 s=2.31×10−4m/s

v3′=4×10−3 m24.0 s=1.67×10−4m/sv4′=4×10−3 m11.4 s =3.51×10−4m/s

v5′=4×10−3 m7.54 s=5.31×10−4m/s

The corresponding charges can be written as,

q1=(6.97×10−19C)(2.52+1.11)2.52=10.0×10−19Cq2=13.4×10−19 C

q3=11.6×10−19Cq4=16.7×10−19Cq5=21.6×10−19C

Therefore, It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Answer 14

(b)

To determine

The charge on each drop and show that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Answer 15

(b)

Expert Solution

Answer to Problem 6P

It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The charge on the droplet is given by,

q1=mgE(v+v1′)v (V)

Here, v is the velocity of fall, v′ is the velocity of rise, E is the electric field, v′ is the upward terminal speed in the presence of the electric field, v is the downward terminal velocity in the absence of the electric field.

The expression for the electric field is given by,

E=Vd (VI)

Conclusion:

Using the times given, the various v′ can be calculated as,

v1′=4×10−3 m36.0 s=1.11×10−4m/sv2′=4×10−3m17.3 s=2.31×10−4m/s

v3′=4×10−3 m24.0 s=1.67×10−4m/sv4′=4×10−3 m11.4 s =3.51×10−4m/s

v5′=4×10−3 m7.54 s=5.31×10−4m/s

The corresponding charges can be written as,

q1=(6.97×10−19C)(2.52+1.11)2.52=10.0×10−19Cq2=13.4×10−19 C

q3=11.6×10−19Cq4=16.7×10−19Cq5=21.6×10−19C

Therefore, It is shown that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Answer 20

(c)

To determine

The electronic charge from the data.

(c)

Expert Solution

Answer to Problem 6P

The average value is 1.67×10−19 C_.

Explanation of Solution

To find an integer n1 such that 1.5×10−19<qn1<2.0×10−19

q16=1.67×10−19Cq28=1.68×10−19Cq37=1.66×10−19Cq410=1.67×10−19C

q513=1.66×10−19C

The amount of charge gained or lost is,

q2−q1=3.4×10−19Cq3−q1=1.6×10−19Cq2−q3=1.8×10−19Cq4−q1=6.7×10−19C

q4−q3=5.1×10−19Cq4−q2=3.4×10−19Cq5−q1=11.6×10−19Cq5−q4=4.9×10−19C

q5−q4=4.9×10−19Cq5−q3=10.0×10−19C

The integers that yield a value of e between 1.5 and 2.0×10−19 C.

q2−q12=1.70×10−19Cq3−q11=1.60×10−19Cq2−q31=1.80×10−19Cq4−q14=1.68×10−19C

q4−q33=1.70×10−19Cq4−q22=1.65×10−19Cq5−q17=1.66×10−19Cq5−q43=1.63×10−19C

q5−q36=1.67×10−19C

The average of all values will be 1.67×10−19 C.

Conclusion:

Therefore, The average value is 1.67×10−19 C_.

Answer 21

(c)

To determine

The electronic charge from the data.

Answer 22

(c)

Expert Solution

Answer to Problem 6P

The average value is 1.67×10−19 C_.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

To find an integer n1 such that 1.5×10−19<qn1<2.0×10−19

q16=1.67×10−19Cq28=1.68×10−19Cq37=1.66×10−19Cq410=1.67×10−19C

q513=1.66×10−19C

The amount of charge gained or lost is,

q2−q1=3.4×10−19Cq3−q1=1.6×10−19Cq2−q3=1.8×10−19Cq4−q1=6.7×10−19C

q4−q3=5.1×10−19Cq4−q2=3.4×10−19Cq5−q1=11.6×10−19Cq5−q4=4.9×10−19C

q5−q4=4.9×10−19Cq5−q3=10.0×10−19C

The integers that yield a value of e between 1.5 and 2.0×10−19 C.

q2−q12=1.70×10−19Cq3−q11=1.60×10−19Cq2−q31=1.80×10−19Cq4−q14=1.68×10−19C

q4−q33=1.70×10−19Cq4−q22=1.65×10−19Cq5−q17=1.66×10−19Cq5−q43=1.63×10−19C

q5−q36=1.67×10−19C

The average of all values will be 1.67×10−19 C.

Conclusion:

Therefore, The average value is 1.67×10−19 C_.

Answer 27

Ch. 4.2 - Exercise 1 Find the horizontal speed vx for this...Ch. 4.2 - Prob. 2E Ch. 4.3 - Prob. 3E Ch. 4.3 - Prob. 4E Ch. 4.3 - Prob. 5E Ch. 4 - Prob. 1Q Ch. 4 - Prob. 2Q Ch. 4 - Prob. 3Q Ch. 4 - Prob. 4Q Ch. 4 - Prob. 5Q

The radius and the mass of the drop used in the experiment.

Concept explainers

The radius and the mass of the drop used in the experiment.

Answer to Problem 6P

Explanation of Solution

The charge on each drop and show that the charge is quantized by considering both the size of each charge and the amount of charge gained when the rise time changes.

Answer to Problem 6P

Explanation of Solution

The electronic charge from the data.

Answer to Problem 6P

Explanation of Solution

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