measuring instrument of potential difference of car accumulator traditional and modern Parts 1 AMNIMARJESLOW GOVERNMENT 91220017 XI XA PIN PING HUNG CHOP LJBUSAF

         Measuring instrument of potential difference of traditional and modern car accumulator   
 
Almost all electronics equipment uses equipment with DC energy source or direct current, especially the components that are related to the process in simple circuit and electronics complex with diode, transistor and IC components, chip, micro controller, microprocessor different from the circuit composed by resistor, inductor and capacitors because everything can be powered by AC and DC current, here we will discuss measuring instrument of different potential of accumulator of car where accumulator is DC current source which supply energy for car starter and other car accessories such as radio, lamp, speaker and so on . 


Electric Potential Energy: Potential Difference 

 will be able to:

• Define electric potential and electric potential energy.
• Describe the relationship between potential difference and electrical potential energy.
• Explain electron volt and its usage in submicroscopic process.
• Determine electric potential energy given potential difference and amount of charge.

Figure 1. A charge accelerated by an electric field is analogous to a mass going down a hill. In both cases potential energy is converted to another form. Work is done by a force, but since this force is conservative, we can write W = –ΔPE.

When a free positive charge q is accelerated by an electric field, such as shown in Figure 1, it is given kinetic energy. The process is analogous to an object being accelerated by a gravitational field. It is as if the charge is going down an electrical hill where its electric potential energy is converted to kinetic energy. Let us explore the work done on a charge q by the electric field in this process, so that we may develop a definition of electric potential energy.

The electrostatic or Coulomb force is conservative, which means that the work done on q is independent of the path taken. This is exactly analogous to the gravitational force in the absence of dissipative forces such as friction. When a force is conservative, it is possible to define a potential energy associated with the force, and it is usually easier to deal with the potential energy (because it depends only on position) than to calculate the work directly.

We use the letters PE to denote electric potential energy, which has units of joules (J). The change in potential energy, ΔPE, is crucial, since the work done by a conservative force is the negative of the change in potential energy; that is, W = –ΔPE. For example, work W done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative ΔPE. There must be a minus sign in front of ΔPE to make W positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

POTENTIAL ENERGY

W = –ΔPE. For example, work W done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative ΔPE. There must be a minus sign in front of ΔPE to make W positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

Gravitational potential energy and electric potential energy are quite analogous. Potential energy accounts for work done by a conservative force and gives added insight regarding energy and energy transformation without the necessity of dealing with the force directly. It is much more common, for example, to use the concept of voltage (related to electric potential energy) than to deal with the Coulomb force directly.

Calculating the work directly is generally difficult, since W = Fd cos θ and the direction and magnitude of F can be complex for multiple charges, for odd-shaped objects, and along arbitrary paths. But we do know that, since F = qE, the work, and hence ΔPE, is proportional to the test charge q. To have a physical quantity that is independent of test charge, we define electric potential V (or simply potential, since electric is understood) to be the potential energy per unit charge $V=\frac{\text{PE}}{q}$.

ELECTRIC POTENTIAL

This is the electric potential energy per unit charge.

$V=\frac{\text{PE}}{q}$

Since PE is proportional to q , the dependence on q cancels. Thus V does not depend on q. The change in potential energy ΔPE is crucial, and so we are concerned with the difference in potential or potential difference ΔV between two points, where

$\Delta V=V_{\text{B}}-V_{\text{A}}=\frac{\Delta \text{PE}}{q}$

The potential difference between points A and B, VB − VA, is thus defined to be the change in potential energy of a charge q moved from A to B, divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta.

$1\text{V}=1\frac{\text{J}}{\text{C}}$

POTENTIAL DIFFERENCE

The potential difference between points A and B, VB – VA, is defined to be the change in potential energy of a charge q moved from A to B, divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta.

$1\text{V}=1\frac{\text{J}}{\text{C}}$

The familiar term voltage is the common name for potential difference. Keep in mind that whenever a voltage is quoted, it is understood to be the potential difference between two points. For example, every battery has two terminals, and its voltage is the potential difference between them. More fundamentally, the point you choose to be zero volts is arbitrary. This is analogous to the fact that gravitational potential energy has an arbitrary zero, such as sea level or perhaps a lecture hall floor.

In summary, the relationship between potential difference (or voltage) and electrical potential energy is given by $\Delta V=\frac{\Delta \text{PE}}{q}$ and ΔPE = qΔV.

POTENTIAL DIFFERENCE AND ELECTRICAL POTENTIAL ENERGY

The relationship between potential difference (or voltage) and electrical potential energy is given by

$\Delta V=\frac{\Delta \text{PE}}{q}$ and ΔPE = qΔV

The second equation is equivalent to the first.

Voltage is not the same as energy. Voltage is the energy per unit charge. Thus a motorcycle battery and a car battery can both have the same voltage (more precisely, the same potential difference between battery terminals), yet one stores much more energy than the other since ΔPE = qΔV. The car battery can move more charge than the motorcycle battery, although both are 12 V batteries.

EXAMPLE 1. CALCULATING ENERGY

Suppose you have a 12.0 V motorcycle battery that can move 5000 C of charge, and a 12.0 V car battery that can move 60,000 C of charge. How much energy does each deliver? (Assume that the numerical value of each charge is accurate to three significant figures.)

Strategy

To say we have a 12.0 V battery means that its terminals have a 12.0 V potential difference. When such a battery moves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to ΔPE = qΔV.

So to find the energy output, we multiply the charge moved by the potential difference.

Solution

For the motorcycle battery, q = 5000 C and ΔV = 12.0 V. The total energy delivered by the motorcycle battery is

$\begin{array}{ccc}\Delta {\text{PE}}_{\text{cycle}}& =& \left(5000\text{C}\right)\left(12.0\text{V}\right)\\ & =& \left(5000\text{C}\right)\left(12.0\text{J/C}\right)\\ & =& 6.00\times {10}^4\text{J}\end{array}$

Similarly, for the car battery, q = 60,000 C and

$\begin{array}{ccc}\Delta {\text{PE}}_{\text{car}}& =& \left(60,000\text{C}\right)\left(12.0\text{V}\right)\\ & =& 7.20\times {10}^5\text{J}\end{array}$

Discussion

While voltage and energy are related, they are not the same thing. The voltages of the batteries are identical, but the energy supplied by each is quite different. Note also that as a battery is discharged, some of its energy is used internally and its terminal voltage drops, such as when headlights dim because of a low car battery. The energy supplied by the battery is still calculated as in this example, but not all of the energy is available for external use.

Note that the energies calculated in the previous example are absolute values. The change in potential energy for the battery is negative, since it loses energy. These batteries, like many electrical systems, actually move negative charge—electrons in particular. The batteries repel electrons from their negative terminals (A) through whatever circuitry is involved and attract them to their positive terminals (B) as shown in Figure 2. The change in potential is ΔV = VB – VA = +12 V and the charge q is negative, so that ΔPE = qΔV is negative, meaning the potential energy of the battery has decreased when q has moved from A to B.

Figure 2. A battery moves negative charge from its negative terminal through a headlight to its positive terminal. Appropriate combinations of chemicals in the battery separate charges so that the negative terminal has an excess of negative charge, which is repelled by it and attracted to the excess positive charge on the other terminal. In terms of potential, the positive terminal is at a higher voltage than the negative. Inside the battery, both positive and negative charges move.

EXAMPLE 2. HOW MANY ELECTRONS MOVE THROUGH A HEADLIGHT EACH SECOND?

When a 12.0 V car battery runs a single 30.0 W headlight, how many electrons pass through it each second?

Strategy

To find the number of electrons, we must first find the charge that moved in 1.00 s. The charge moved is related to voltage and energy through the equation ΔPE =  qΔV. A 30.0 W lamp uses 30.0 joules per second. Since the battery loses energy, we have ΔPE = –30.0 J and, since the electrons are going from the negative terminal to the positive, we see that ΔV = +12.0V.

Solution

To find the charge q moved, we solve the equation ΔPE = qΔV: $q=\frac{\Delta \text{PE}}{\Delta V}$.

Entering the values for ΔPE and ΔV, we get

$q=\frac{-30.0\text{J}}{+12.0\text{V}}=\frac{-30.0\text{J}}{+12.0\text{J/C}}-2.50\text{C}$

The number of electrons ne is the total charge divided by the charge per electron. That is,

${\text{n}}_{\text{e}}=\frac{-2.50\text{C}}{-1.60\times {10}^{-19}{\text{C/e}}^{-}}=1.56\times {10}^{19}\text{electrons}$

Discussion

This is a very large number. It is no wonder that we do not ordinarily observe individual electrons with so many being present in ordinary systems. In fact, electricity had been in use for many decades before it was determined that the moving charges in many circumstances were negative. Positive charge moving in the opposite direction of negative charge often produces identical effects; this makes it difficult to determine which is moving or whether both are moving.

The Electron Volt

Figure 3. A typical electron gun accelerates electrons using a potential difference between two metal plates. The energy of the electron in electron volts is numerically the same as the voltage between the plates. For example, a 5000 V potential difference produces 5000 eV electrons.

The energy per electron is very small in macroscopic situations like that in the previous example—a tiny fraction of a joule. But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. For example, even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue. The particle may do its damage by direct collision, or it may create harmful x rays, which can also inflict damage. It is useful to have an energy unit related to submicroscopic effects. Figure 3 shows a situation related to the definition of such an energy unit. An electron is accelerated between two charged metal plates as it might be in an old-model television tube or oscilloscope. The electron is given kinetic energy that is later converted to another form—light in the television tube, for example. (Note that downhill for the electron is uphill for a positive charge.) Since energy is related to voltage by ΔPE = qΔV, we can think of the joule as a coulomb-volt.

On the submicroscopic scale, it is more convenient to define an energy unit called the electron volt(eV), which is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,

$\begin{array}{ccc}1\text{eV}& =& \left(1.60\times {10}^{-19}\text{C}\right)\left(1\text{V}\right)=\left(1.60\times {10}^{-19}\text{C}\right)\left(1\text{J/C}\right)\\ & =& 1.60\times {10}^{-19}\text{J}\end{array}$

ELECTRON VOLT

On the submicroscopic scale, it is more convenient to define an energy unit called the electron volt (eV), which is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,

$\begin{array}{ccc}1\text{eV}& =& \left(1.60\times {10}^{-19}\text{C}\right)\left(1\text{V}\right)=\left(1.60\times {10}^{-19}\text{C}\right)\left(1\text{J/C}\right)\\ & =& 1.60\times {10}^{-19}\text{J}\end{array}$

An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. It follows that an electron accelerated through 50 V is given 50 eV. A potential difference of 100,000 V (100 kV) will give an electron an energy of 100,000 eV (100 keV), and so on. Similarly, an ion with a double positive charge accelerated through 100 V will be given 200 eV of energy. These simple relationships between accelerating voltage and particle charges make the electron volt a simple and convenient energy unit in such circumstances.

MAKING CONNECTIONS: ENERGY UNITS

The electron volt (eV) is the most common energy unit for submicroscopic processes. This will be particularly noticeable in the chapters on modern physics. Energy is so important to so many subjects that there is a tendency to define a special energy unit for each major topic. There are, for example, calories for food energy, kilowatt-hours for electrical energy, and therms for natural gas energy.

The electron volt is commonly employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies are among the quantities often expressed in electron volts. For example, about 5 eV of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of 30 kV, it is given an energy of 30 keV (30,000 eV) and it can break up as many as 6000 of these molecules (30,000 eV ÷ 5 eV per molecule= 6000 molecules). Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and can, thus, produce significant biological damage.

Conservation of Energy

The total energy of a system is conserved if there is no net addition (or subtraction) of work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.

Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, KE+PE = constant. A loss of PE of a charged particle becomes an increase in its KE. Here PE is the electric potential energy. Conservation of energy is stated in equation form as KE + PE = constant or KEi + PE i = KEf + PEf, where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.

EXAMPLE 3. ELECTRICAL POTENTIAL ENERGY CONVERTED TO KINETIC ENERGY

Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value is accurate to three significant figures.)

Strategy

We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (we will check on this assumption later), all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms of energy to be KEi = 0, $K{E}_f=\frac{1}{2}m{v}^2$, PEi = qV, and PEf = 0.

Solution

Conservation of energy states that KEi + PE i = KE f + PE f .

Entering the forms identified above, we obtain $qV=\frac{m{v}^2}{2}$.

We solve this for v:

$v=\sqrt{\frac{2qV}{m}}$

Entering values for q, V, and m gives

$\begin{array}{ccc}v& =& \sqrt{\frac{2\left(-1.60\times {10}^{-19}\text{C}\right)\left(-100\text{J/C}\right)}{9.11\times {10}^{-31}\text{kg}}}\\ & =& 5.93\times {10}^6\text{m/s}\end{array}$

Discussion

Note that both the charge and the initial voltage are negative, as in Figure 3. From the discussions in Electric Charge and Electric Field, we know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. Those higher voltages produce electron speeds so great that relativistic effects must be taken into account. That is why a low voltage is considered (accurately) in this example.

Section Summary

• Electric potential is potential energy per unit charge.
• The potential difference between points A and B, VB − VA, defined to be the change in potential energy of a charge q moved from A to B, is equal to the change in potential energy divided by the charge, Potential difference is commonly called voltage, represented by the symbol ΔV: $\Delta V=\frac{\Delta \text{PE}}{q}$ and ΔPE = qΔV.
• An electron volt is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,
  $\begin{array}{ccc}\text{1 eV}& =& \left(1.60\times {\text{10}}^{\text{-19}}\text{C}\right)\left(1V\right)=\left(1.60\times {\text{10}}^{\text{-19}}\text{C}\right)\left(1J/C\right)\\ & =& 1.60\times {\text{10}}^{\text{-19}}\text{J.}\end{array}$
• Mechanical energy is the sum of the kinetic energy and potential energy of a system, that is, KE + PE. This sum is a constant.

CONCEPTUAL QUESTIONS

1. Voltage is the common word for potential difference. Which term is more descriptive, voltage or potential difference?
2. If the voltage between two points is zero, can a test charge be moved between them with zero net work being done? Can this necessarily be done without exerting a force? Explain.
3. What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential energy?
4. Voltages are always measured between two points. Why?
5. How are units of volts and electron volts related? How do they differ?

PROBLEMS & EXERCISES

1. Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be 1.67 × 10−27kg.
2. An evacuated tube uses an accelerating voltage of 40 kV to accelerate electrons to hit a copper plate and produce x rays. Non-relativistically, what would be the maximum speed of these electrons?
3. A bare helium nucleus has two positive charges and a mass of 6.64 × 10−27 kg. (a) Calculate its kinetic energy in joules at 2.00% of the speed of light. (b) What is this in electron volts? (c) What voltage would be needed to obtain this energy?
4. Integrated Concepts. Singly charged gas ions are accelerated from rest through a voltage of 13.0 V. At what temperature will the average kinetic energy of gas molecules be the same as that given these ions?
5. Integrated Concepts. The temperature near the center of the Sun is thought to be 15 million degrees Celsius (1.5 × 107 ºC). Through what voltage must a singly charged ion be accelerated to have the same energy as the average kinetic energy of ions at this temperature?
6. Integrated Concepts. (a) What is the average power output of a heart defibrillator that dissipates 400 J of energy in 10.0 ms? (b) Considering the high-power output, why doesn’t the defibrillator produce serious burns?
7. Integrated Concepts. A lightning bolt strikes a tree, moving 20.0 C of charge through a potential difference of 1.00 × 102 MV. (a) What energy was dissipated? (b) What mass of water could be raised from 15ºC to the boiling point and then boiled by this energy? (c) Discuss the damage that could be caused to the tree by the expansion of the boiling steam.
8. Integrated Concepts. A 12.0 V battery-operated bottle warmer heats 50.0 g of glass, 2.50 × 102 g of baby formula, and 2.00 × 102 g of aluminum from 20.0ºC to 90.0ºC. (a) How much charge is moved by the battery? (b) How many electrons per second flow if it takes 5.00 min to warm the formula? (Hint: Assume that the specific heat of baby formula is about the same as the specific heat of water.)
9. Integrated Concepts. A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00 × 102 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00 × 102 N force for an hour.
10. Integrated Concepts. Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. (a) Calculate the potential energy of two singly charged nuclei separated by 1.00 × 10−12 m by finding the voltage of one at that distance and multiplying by the charge of the other. (b) At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?
11. Unreasonable Results. (a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00 C of excess positive charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible?
12. Construct Your Own Problem. Consider a battery used to supply energy to a cellular phone. Construct a problem in which you determine the energy that must be supplied by the battery, and then calculate the amount of charge it must be able to move in order to supply this energy. Among the things to be considered are the energy needs and battery voltage. You may need to look ahead to interpret manufacturer’s battery ratings in ampere-hours as energy in joules.
13. electric potential: potential energy per unit charge . potential difference (or voltage): change in potential energy of a charge moved from one point to another, divided by the charge; units of potential difference are joules per coulomb, known as volt . electron volt: the energy given to a fundamental charge accelerated through a potential difference of one volt mechanical energy: sum of the kinetic energy and potential energy of a system; this sum is a constant

SELECTED SOLUTIONS TO PROBLEMS & EXERCISES

1. 42.8

4. 1.00 × 105 K

6. (a) 4 × 104 W; (b) A defibrillator does not cause serious burns because the skin conducts electricity well at high voltages, like those used in defibrillators. The gel used aids in the transfer of energy to the body, and the skin doesn’t absorb the energy, but rather lets it pass through to the heart.

8. (a) 7.40 × 103 C; (b) 1.54 × 1020 electrons per second

9. 3.89 × 106 C

11. (a) 1.44 × 1012 V; (b) This voltage is very high. A 10.0 cm diameter sphere could never maintain this voltage; it would discharge; (c) An 8.00 C charge is more charge than can reasonably be accumulated on a sphere of that size. 

Electric Potential in a Uniform Electric Field 

section, we will explore the relationship between voltage and electric field. For example, a uniform electric field E is produced by placing a potential difference (or voltage) ΔV across two parallel metal plates, labeled A and B. (See Figure 1.)

Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal a more fundamental relationship between electric potential and electric field. From a physicist’s point of view, either ΔV or E can be used to describe any charge distribution. ΔV is most closely tied to energy, whereas E is most closely related to force. ΔV is a scalar quantity and has no direction, while E is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by E below.) The relationship between ΔV and E is revealed by calculating the work done by the force in moving a charge from point A to point B.

But, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electric field as an interesting special case.

The work done by the electric field in Figure 1 to move a positive charge q from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

W = −ΔPE = −qΔV.

The potential difference between points A and B is

−ΔV = −(VB − VA) = VA − VB = VAB.

Entering this into the expression for work yields W = qVAB.

Work is W = Fd cos θ; here cos θ = 1, since the path is parallel to the field, and so W = Fd. Since F = qE, we see that W = qEd. Substituting this expression for work into the previous equation gives qEd = qVAB.

The charge cancels, and so the voltage between points A and B is seen to be

$\{\begin{array}{ccc}{V}_{\text{AB}}& =& Ed\\ E& =& \frac{V_{\text{AB}}}{d}\end{array}$ (uniform E − field only)

where d is the distance from A to B, or the distance between the plates in Figure 1. Note that the above equation implies the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus the following relation among units is valid: 1 N/C = 1 V/m.

VOLTAGE BETWEEN POINTS A AND B

$\{\begin{array}{ccc}{V}_{\text{AB}}& =& Ed\\ E& =& \frac{V_{\text{AB}}}{d}\end{array}$ (uniform E − field only)

where d is the distance from A to B, or the distance between the plates.

EXAMPLE 1. WHAT IS THE HIGHEST VOLTAGE POSSIBLE BETWEEN TWO PLATES?

Dry air will support a maximum electric field strength of about 3.0 × 106 V/m. Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air?

Strategy

We are given the maximum electric field E between the plates and the distance d between them. The equation VAB = Ed can thus be used to calculate the maximum voltage.

Solution

The potential difference or voltage between the plates is

VAB = Ed.

Entering the given values for E and d gives

VAB = (3.0 × 106 V/m)(0.025 m) 7.5 × 104 V or VAB = 75 kV.

(The answer is quoted to only two digits, since the maximum field strength is approximate.)

Discussion

One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5 cm (1 in.) gap, or 150 kV for a 5 cm spark. This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage will cause a spark if there are points on the surface, since points create greater fields than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up, say with static electricity, on dry days.

Figure 2. A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles as they pass through the gas between the plates allows a spark to jump. The sparks are perpendicular to the plates, following electric field lines between them. The potential difference between adjacent plates is not high enough to cause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). (credit: Daderot, Wikimedia Commons)

EXAMPLE 2. FIELD AND FORCE INSIDE AN ELECTRON GUN

1. An electron gun has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. What is the electric field strength between the plates?
2. What force would this field exert on a piece of plastic with a 0.500 μC charge that gets between the plates?

Strategy

Since the voltage and plate separation are given, the electric field strength can be calculated directly from the expression $E=\frac{V_{\text{AB}}}{d}$. Once the electric field strength is known, the force on a charge is found using F = qE. Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, F = qE.

Solution for Part 1

The expression for the magnitude of the electric field between two uniform metal plates is

$E=\frac{V_{\text{AB}}}{d}$.

Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0 kV. Entering this value for VAB and the plate separation of 0.0400 m, we obtain

$E=\frac{25.0\text{kV}}{0.0400\text{m}}=6.25\times {10}^5\text{V/m}$

Solution for Part 2

The magnitude of the force on a charge in an electric field is obtained from the equation F = qE.

Substituting known values gives

F = (0.500 × 10−6 C)(6.25 × 105 V/m) = 0.313 N.

Discussion

Note that the units are newtons, since 1 V/m = 1 N/C. The force on the charge is the same no matter where the charge is located between the plates. This is because the electric field is uniform between the plates.

In more general situations, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because the force on a positive charge is in the direction of E and also in the direction of lower potential V. Furthermore, the magnitude of E equals the rate of decrease of V with distance. The faster V decreases over distance, the greater the electric field. In equation form, the general relationship between voltage and electric field is

$E=-\frac{\Delta V}{\Delta s}$,

where Δs is the distance over which the change in potential, ΔV, takes place. The minus sign tells us that E points in the direction of decreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electric potential.

RELATIONSHIP BETWEEN VOLTAGE AND ELECTRIC FIELD

In equation form, the general relationship between voltage and electric field is

$E=-\frac{\Delta V}{\Delta s}$,

where Δs is the distance over which the change in potential, ΔV, takes place. The minus sign tells us that E points in the direction of decreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electric potential.

For continually changing potentials, ΔV and Δs become infinitesimals and differential calculus must be employed to determine the electric field.

Section Summary

• The voltage between points A and B is

$\{\begin{array}{ccc}{V}_{\text{AB}}& =& Ed\\ E& =& \frac{V_{\text{AB}}}{d}\end{array}$ (uniform E − field only)

where d is the distance from A to B, or the distance between the plates.

• In equation form, the general relationship between voltage and electric field is

$E=-\frac{\Delta V}{\Delta s}$,

where Δs is the distance over which the change in potential, ΔV, takes place. The minus sign tells us that E points in the direction of decreasing potential.) The electric field is said to be the gradient (as in grade or slope) of the electric potential.

CONCEPTUAL QUESTIONS

1. Discuss how potential difference and electric field strength are related. Give an example.
2. What is the strength of the electric field in a region where the electric potential is constant?
3. Will a negative charge, initially at rest, move toward higher or lower potential? Explain why.

PROBLEMS & EXERCISES

1. Show that units of V/m and N/C for electric field strength are indeed equivalent.
2. What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50 × 104 V?
3. The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50 × 104 V/m. (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?
4. How far apart are two conducting plates that have an electric field strength of 4.50× 103 V/m between them, if their potential difference is 15.0 kV?
5. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air (3.0 × 106 V/m) if the plates are separated by 2.00 mm and a potential difference of 5.0 × 103 V is applied? (b) How close together can the plates be with this applied voltage?
6. The voltage across a membrane forming a cell wall is 80.0 mV and the membrane is 9.00 nm thick. What is the electric field strength? (The value is surprisingly large, but correct. Membranes are discussed in Capacitors and Dielectrics and Nerve Conduction—Electrocardiograms.) You may assume a uniform electric field.
7. Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. (Membranes are discussed in some detail in Nerve Conduction—Electrocardiograms.) What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m? You may assume a uniform electric field.
8. Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate (and 2.00 cm from the other) is 450 V? (b) What is the voltage between the plates?
9. Find the maximum potential difference between two parallel conducting plates separated by 0.500 cm of air, given the maximum sustainable electric field strength in air to be 3.0 × 106 V/m.
10. A doubly charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the plates?
11. An electron is to be accelerated in a uniform electric field having a strength of 2.00 × 106 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?

Glossary

scalar: physical quantity with magnitude but no direction

vector: physical quantity with both magnitude and direction

SELECTED SOLUTIONS TO PROBLEMS & EXERCISES

3. (a) 3.00 kV; (b) 750 V

5. (a) No. The electric field strength between the plates is 2.5 × 106 V/m, which is lower than the breakdown strength for air (3.0 × 106 V/m}); (b) 1.7 mm

7. 44.0 mV

9. 15 kV

11. (a) 800 KeV; (b) 25.0 km 

Electrical Potential Due to a Point Charge 

Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (like on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. Using calculus to find the work needed to move a test charge q from a large distance away to a distance of r from a point charge Q, and noting the connection between work and potential (W = −qΔV), it can be shown that the electric potential V of a point charge is $V=\frac{kQ}{r}$ (Point Charge), where k is a constant equal to 9.0 × 109 N · m2/C2.

ELECTRIC POTENTIAL V OF A POINT CHARGE

The electric potential V of a point charge is given by

$V=\frac{kQ}{r}$ (Point Charge)

The potential at infinity is chosen to be zero. Thus V for a point charge decreases with distance, whereas E for a point charge decreases with distance squared:

$E=\frac{F}{q}=\frac{kQ}{r^2}$.

Recall that the electric potential V is a scalar and has no direction, whereas the electric field E is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact that V is closely associated with energy, a scalar, whereas E is closely associated with force, a vector.

EXAMPLE 1. WHAT VOLTAGE IS PRODUCED BY A SMALL CHARGE ON A METAL SPHERE?

Charges in static electricity are typically in the nanocoulomb (nC) to microcoulomb (µC) range. What is the voltage 5.00 cm away from the center of a 1-cm diameter metal sphere that has a −3.00 nC static charge?

Strategy

As we have discussed in Electric Charge and Electric Field, charge on a metal sphere spreads out uniformly and produces a field like that of a point charge located at its center. Thus we can find the voltage using the equation $V=k\frac{Q}{r}$.

Solution

Entering known values into the expression for the potential of a point charge, we obtain

$\begin{array}{ccc}V& =& k\frac{Q}{r}\\ & =& \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(\frac{-3.00\times {10}^{-9}\text{C}}{5.00\times {10}^{-2}\text{m}}\right)\\ & =& -539\text{V}\end{array}$

Discussion

The negative value for voltage means a positive charge would be attracted from a larger distance, since the potential is lower (more negative) than at larger distances. Conversely, a negative charge would be repelled, as expected.

EXAMPLE 2. WHAT IS THE EXCESS CHARGE ON A VAN DE GRAAFF GENERATOR

Figure 1. The voltage of this demonstration Van de Graaff generator is measured between the charged sphere and ground. Earth’s potential is taken to be zero as a reference. The potential of the charged conducting sphere is the same as that of an equal point charge at its center.

A demonstration Van de Graaff generator has a 25.0 cm diameter metal sphere that produces a voltage of 100 kV near its surface. (See Figure 1.) What excess charge resides on the sphere? (Assume that each numerical value here is shown with three significant figures.)

Strategy

The potential on the surface will be the same as that of a point charge at the center of the sphere, 12.5 cm away. (The radius of the sphere is 12.5 cm.) We can thus determine the excess charge using the equation $V=\frac{kQ}{r}$.

Solution

Solving for Q and entering known values gives

$\begin{array}{ccc}Q& =& \frac{rV}{k}\\ & =& \frac{\left(0.125\text{m}\right)\left(100\times {10}^3\text{V}\right)}{8.99\times {10}^9\text{N}\cdot {\text{m}}^2{\text{/C}}^2}\\ & =& 1.39\times {10}^{-6}\text{C}=1.39\mu \text{C}\end{array}$

Discussion

This is a relatively small charge, but it produces a rather large voltage. We have another indication here that it is difficult to store isolated charges.

The voltages in both of these examples could be measured with a meter that compares the measured potential with ground potential. Ground potential is often taken to be zero (instead of taking the potential at infinity to be zero). It is the potential difference between two points that is of importance, and very often there is a tacit assumption that some reference point, such as Earth or a very distant point, is at zero potential. As noted in Electric Potential Energy: Potential Difference, this is analogous to taking sea level as h = 0 when considering gravitational potential energy, PEg = mgh.

Section Summary

• Electric potential of a point charge is $V=\frac{kQ}{r}$ .
• Electric potential is a scalar, and electric field is a vector. Addition of voltages as numbers gives the voltage due to a combination of point charges, whereas addition of individual fields as vectors gives the total electric field.

CONCEPTUAL QUESTIONS

1. In what region of space is the potential due to a uniformly charged sphere the same as that of a point charge? In what region does it differ from that of a point charge?
2. Can the potential of a non-uniformly charged sphere be the same as that of a point charge? Explain.

PROBLEMS & EXERCISES

1. A 0.500 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 40.0 pC charge on its surface. What is the potential near its surface?
2. What is the potential 0.530 × 10−10 m from a proton (the average distance between the proton and electron in a hydrogen atom)?
3. (a) A sphere has a surface uniformly charged with 1.00 C. At what distance from its center is the potential 5.00 MV? (b) What does your answer imply about the practical aspect of isolating such a large charge?
4. How far from a 1.00 μC point charge will the potential be 100 V? At what distance will it be 2.00 × 102 V?
5. What are the sign and magnitude of a point charge that produces a potential of −2.00 V at a distance of 1.00 mm?
6. If the potential due to a point charge is 5.00 × 102 V at a distance of 15.0 m, what are the sign and magnitude of the charge?
7. In nuclear fission, a nucleus splits roughly in half. (a) What is the potential 2.00 × 10−14 m from a fragment that has 46 protons in it? (b) What is the potential energy in MeV of a similarly charged fragment at this distance?
8. A research Van de Graaff generator has a 2.00-m-diameter metal sphere with a charge of 5.00 mC on it. (a) What is the potential near its surface? (b) At what distance from its center is the potential 1.00 MV? (c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV at this distance?
9. An electrostatic paint sprayer has a 0.200-m-diameter metal sphere at a potential of 25.0 kV that repels paint droplets onto a grounded object. (a) What charge is on the sphere? (b) What charge must a 0.100-mg drop of paint have to arrive at the object with a speed of 10.0 m/s?
10. In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was 5.00 MeV, how close to the gold nucleus (79 protons) could it come before being deflected?
11. (a) What is the potential between two points situated 10 cm and 20 cm from a 3.0 µC point charge? (b) To what location should the point at 20 cm be moved to increase this potential difference by a factor of two?
12. Unreasonable Results. (a) What is the final speed of an electron accelerated from rest through a voltage of 25.0 MV by a negatively charged Van de Graaff terminal? (b) What is unreasonable about this result? (c) Which assumptions are responsible?

SELECTED SOLUTIONS TO PROBLEMS & EXERCISES

1. 144 V

3. (a) 1.80 km; (b) A charge of 1 C is a very large amount of charge; a sphere of radius 1.80 km is not practical.

5. −2.22 × 10−13 C

7. (a) 3.31 × 106 V; (b) 152 MeV

9. (a) 2.78 × 10−7 C; (b) 2.00 × 10−10 C

12. (a) 2.96 × 109 m/s; (b) This velocity is far too great. It is faster than the speed of light; (c) The assumption that the speed of the electron is far less than that of light and that the problem does not require a relativistic treatment produces an answer greater than the speed of light.

Equipotential Lines