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Chapter 29: Problem 17

$$\text { Show that } 1 \mathrm{V} / \mathrm{m}=1 \mathrm{N} / \mathrm{C}$$

Short Answer

Expert verified
Yes, 1 V/m is equivalent to 1 N/C. They both represent the electric field strength or the force per unit charge.

Step by step solution

01

Understanding the basic units.

First off, start with understanding the basic units. The volt (V) is the unit of electric potential, the metre (m) is the unit of distance, the Newton (N) is the unit of force, and the Coulomb (C) is the unit of electric charge. In particular, 1 Newton is defined as the force required to accelerate 1 kg mass at a rate of 1 m/s虏, and 1 Volt is defined as the energy per unit charge, or Joules per Coulomb.
02

Show that 1 V/m is equivalent to 1 N/C.

Converting volt (energy per unit charge) to force, 1 V = 1 J/C. Now, remember that work done (or energy) is equivalent to force x distance (W = Fd), meaning that Joules can be expressed as N.m. Substitute this into the previous equation: 1 V = 1 N.m/C. If you divide both sides by m (because we are interested in force per charge per distance), you get 1 V/m = 1 N/C.
03

Conclusion

Based on the derivations, it can be concluded that 1 V/m is indeed equivalent to 1 N/C. Both of these units describe the electric field strength, i.e., the force experienced by a unit positive charge in the electric field.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential
Electric potential is a fundamental concept in electromagnetism and electrical engineering. It is essentially the energy per unit charge needed to bring a small positive test charge from infinity to a specific point in an electric field. To understand this better, imagine electric potential as a measure of how much work is needed or stored to move charges within an electric field.
  • Measured in Volts (V), it indicates the potential energy difference per charge.
  • A high electric potential means more energy is required to move charges.
When we say 1 V, we are referring to 1 Joule of energy per Coulomb of charge. Therefore, electric potential considers how energy and charge interact in space.
Unit Conversions
Unit conversions play a critical role in understanding and solving physics problems, especially when dealing with complex concepts like electric fields. It's vital to switch seamlessly between different measurement units to interpret the physical world correctly without contradiction.
  • 1 Volt (V) = 1 Joule (J) per Coulomb (C). This defines how much energy each charge unit carries or relinquishes.
  • Joules can also be viewed as Newton meters (N路m), linking energy to force and distance.
In our context of proving that 1 V/m equals 1 N/C, a grasp of these conversions helps establish how energy distribution per unit charge and distance equates to force per unit charge. This makes unit conversions an invaluable tool in bridging different dimensions of physics.
Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. Simply put, charges drive electrical phenomena.
  • Coulombs (C) are the standard unit of electric charge.
  • There are two types 鈥� positive and negative, typically observed as protons and electrons.
Charges interact with electric fields to produce forces, quantified by their product with electric field intensity (N/C). An understanding of electric charge is essential, as it connects to the electric potential, showing how forces manifest in an electric field.
Force and Energy Relations
Understanding how force and energy interrelate helps ina explaining the concepts tied to electric fields. Specifically, it highlights the dynamic between exerted force and energy transformation.
  • Force (measured in Newtons, N) is what moves objects within a field.
  • Energy, viewed as work, is often expressed as force multiplied by distance (W = Fd).
Consequently, the conversion of volts to newtons per coulomb bridges the gap between these quantities, ensuring energy distribution is perceived as actual force exertion over a distance. It's through these relationships that we comprehend transformations in electric fields, expressed in terms that quantify the influence and interaction of charges.

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Most popular questions from this chapter

Two 10 -cm-diameter electrodes \(0.50 \mathrm{cm}\) apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 15 V battery. After a long time, the capacitor is disconnected from the battery but is not discharged. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes a. Right after the battery is disconnected? b. After insulating handles are used to pull the electrodes away from each other until they are \(1.0 \mathrm{cm}\) apart? c. After the original electrodes (not the modified electrodes of part b) are expanded until they are \(20 \mathrm{cm}\) in diameter?

The electric field strength is \(50,000 \mathrm{N} / \mathrm{C}\) inside a parallel plate capacitor with a \(2.0 \mathrm{mm}\) spacing. A proton is released from rest at the positive plate. What is the proton's speed when it reaches the negative plate?

You are given the equation(s) used to solve a problem. For each of these, a. Write a realistic problem for which this is the correct equation(s). b. Finish the solution of the problem. $$\begin{array}{l} \frac{\left(9.0 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right)\left(3.0 \times 10^{-9} \mathrm{C}\right)}{0.030 \mathrm{m}}+ \\ \frac{\left(9.0 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right)\left(3.0 \times 10^{-9} \mathrm{C}\right)}{(0.030 \mathrm{m})+d}=1200 \mathrm{V} \end{array}$$

An arrangement of source charges produces the electric potential \(V=5000 x^{2}\) along the \(x\) -axis, where \(V\) is in volts and \(x\) is in meters. a. Graph the potential between \(x=-10 \mathrm{cm}\) and \(x=+10 \mathrm{cm} .\) b. Describe the motion of a positively charged particle in this potential. c. What is the mechanical energy of a \(1.0 \mathrm{g}, 10\) nC charged particle if its turning points are at \(\pm 8.0 \mathrm{cm} ?\) d. What is the particle's maximum speed?

a. Find an algebraic expression for the electric field strength \(E_{0}\) at the surface of a charged sphere in terms of the sphere's potential \(V_{0}\) and radius \(R\) b. What is the electric field strength at the surface of a \(1.0-\mathrm{cm}-\) diameter marble charged to 500 V?
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