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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/s²). 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

If a capacitor has a capacitance of \(10.2 \mu \mathrm{F}\) and we wish to lower the potential difference across the plates by \(60.0 \mathrm{V},\) what magnitude of charge will we have to remove from each plate?
A charge of \(+2.0 \mathrm{mC}\) is located at \(x=0, y=0\) and a charge of $-4.0 \mathrm{mC}\( is located at \)x=0, y=3.0 \mathrm{m} .$ What is the electric potential due to these charges at a point with coordinates $x=4.0 \mathrm{m}, y=0 ?$
A spherical conductor with a radius of \(75.0 \mathrm{cm}\) has an electric field of magnitude \(8.40 \times 10^{5} \mathrm{V} / \mathrm{m}\) just outside its surface. What is the electric potential just outside the surface, assuming the potential is zero far away from the conductor?
A defibrillator is used to restart a person's heart after it stops beating. Energy is delivered to the heart by discharging a capacitor through the body tissues near the heart. If the capacitance of the defibrillator is $9 \mu \mathrm{F}\( and the energy delivered is to be \)300 \mathrm{J},$ to what potential difference must the capacitor be charged?
Find the electric potential energy for the following array of charges: charge \(q_{1}=+4.0 \mu \mathrm{C}\) is located at \((x, y)=(0.0,0.0) \mathrm{m} ;\) charge \(q_{2}=+3.0 \mu \mathrm{C}\) is located at \((4.0,3.0) \mathrm{m} ;\) and charge \(q_{3}=-1.0 \mu \mathrm{C}\) is located at (0.0,3.0) \(\mathrm{m}\)
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