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Chapter 17: Problem 30

By rewriting each unit in terms of kilograms, meters, seconds, and coulombs, show that \(1 \mathrm{N} / \mathrm{C}=1 \mathrm{V} / \mathrm{m}\)

Short Answer

Expert verified
Question: Prove that \(1 \mathrm{N} / \mathrm{C}=1 \mathrm{V} / \mathrm{m}\) using base SI units. Answer: We showed that the given equation is true when expressed in terms of the base SI units (kilograms, meters, seconds, and coulombs) by converting Newtons, Coulombs, and Volts to their respective base SI units, inserting these into the given equation, and simplifying to show that both sides are equal.

Step by step solution

01

Express given equation in terms of base SI units

To show that \(1 \mathrm{N} / \mathrm{C}=1 \mathrm{V} / \mathrm{m}\), first, we need to express both sides in terms of base SI units. The base SI units in this equation are kilogram (kg), meter (m), second (s), and coulomb (C). Newton (N) can be expressed in terms of kg, m, and s while volt (V) can be expressed in terms of kg, m, s, and C.
02

Convert Newton (N) to base SI units

Newton (N) is the unit of force, which can be expressed as the product of mass (in kg) and acceleration (in m/s虏). The equation for force is \(F = ma\) where F is force, m is mass, and a is acceleration. Therefore, 1 Newton (N) can be expressed as \(1\mathrm{N} = 1\mathrm{kg}\, \mathrm{m}/\mathrm{s}^2\).
03

Convert Coulomb (C) to base SI units

Coulomb (C) is the unit of electric charge, which is already a base SI unit. Therefore, there is no need for further conversion.
04

Convert Volt (V) to base SI units

Volt (V) is the unit of electric potential, which can be expressed as energy (in Joules) per unit charge (in Coulombs). The equation for electric potential is \(V = \frac{E}{q}\) where V is the electric potential, E is the energy, and q is the charge. Since 1 Joule (J) can be expressed as \(1 \mathrm{J} = 1 \mathrm{kg} \, \mathrm{m}^2 / \mathrm{s}^2\), 1 Volt (V) can be expressed as \(1 \mathrm{V} = 1 \mathrm{kg} \, \mathrm{m}^2 / (\mathrm{s}^2 \, \mathrm{C})\).
05

Insert the base SI units into the given equation

Now, we will substitute the base SI units for Newton (N), Coulomb (C), and Volt (V) in the given equation, \(1 \mathrm{N} / \mathrm{C}=1 \mathrm{V} / \mathrm{m}\): \(1\mathrm{kg}\, \mathrm{m}/\mathrm{s}^2/\mathrm{C} = 1\mathrm{kg} \, \mathrm{m}^2 / (\mathrm{s}^2 \, \mathrm{C}) / \mathrm{m}\).
06

Simplify the equation

Now, simplify the equation by cancelling out the common terms: \(\frac{1\mathrm{kg}\,\mathrm{m}/\mathrm{s}^2/\mathrm{C}}{1} = \frac{1\mathrm{kg}\, \mathrm{m}^2 / (\mathrm{s}^2 \, \mathrm{C})}{\mathrm{m}}\). \(\frac{1\mathrm{kg}\,\mathrm{m}/\mathrm{s}^2}{\mathrm{C}} = \frac{1\mathrm{kg}\,\mathrm{m}/\mathrm{s}^2}{\mathrm{C}}\). Hence, we have shown that \(1 \mathrm{N} / \mathrm{C}=1 \mathrm{V} / \mathrm{m}\) is true when the units are expressed in terms of their base SI units (kilograms, meters, seconds, and coulombs).

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Most popular questions from this chapter

A parallel plate capacitor is charged by connecting it to a \(12-V\) battery. The battery is then disconnected from the capacitor. The plates are then pulled apart so the spacing between the plates is increased. What is the effect (a) on the electric field between the plates? (b) on the potential difference between the plates?
(a) Calculate the capacitance per unit length of an axon of radius $5.0 \mu \mathrm{m} \text { (see Fig. } 17.14) .$ The membrane acts as an insulator between the conducting fluids inside and outside the neuron. The membrane is 6.0 nm thick and has a dielectric constant of \(7.0 .\) (Note: The membrane is thin compared with the radius of the axon, so the axon can be treated as a parallel plate capacitor.) (b) In its resting state (no signal being transmitted), the potential of the fluid inside is about \(85 \mathrm{mV}\) lower than the outside. Therefore, there must be small net charges \(\pm Q\) on either side of the membrane. Which side has positive charge? What is the magnitude of the charge density on the surfaces of the membrane?
A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
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A parallel plate capacitor is attached to a battery that supplies a constant voltage. While the battery is still attached, a dielectric of dielectric constant \(\kappa=3.0\) is inserted so that it just fits between the plates. What is the energy stored in the capacitor after the dielectric is inserted in terms of the energy \(U_{0}\) before the dielectric was inserted?
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