Electric Field Between Two Parallel Plates
The electric field between two parallel plates is a uniform field that exists between the plates when a potential difference is applied across them. The field lines are straight and parallel, with the direction of the field pointing from the positive plate to the negative plate. The magnitude of the electric field is determined by the potential difference and the distance between the plates.
6 Key excerpts on "Electric Field Between Two Parallel Plates"
eBook - ePub- W. Bolton(Author)
- 2015(Publication Date)
- Routledge(Publisher)
...away from another positive charge or towards a negative charge, while a negative charge in an electric field will move in the opposite direction. We can define the electric field strength E as the force F experienced per unit charge placed in a field and so for a charge q : 13.2 Capacitor If a pair of parallel plates separated by an insulator, e.g. air, are connected to a direct voltage supply such as a battery (Figure 13.2), the battery pushes electrons through the circuit onto one of the plates and removes them from the other plate. The result is that the plate that has gained electrons becomes negatively charged and the plate that has lost electrons becomes positively charged. The result is an electric field between the plates. The electric field is, with the exception of near the edges of the plates, constant between the plates and at right angles to them. In this situation there is a potential difference between the plates. Figure 13.2 (a) Charging a pair of parallel plates, (b) field pattern between parallel oppositely charged plates The potential difference V between the plates tells us the energy used to move a charge between the plates; energy = Vq (see Chapter 9). The work done must be the product of the force F acting on the charge and the distance d moved, i.e. the plate separation. Thus: Fd = Vq Since we have defined the electric field strength E as F/q, then: The unit of electric field strength is thus either newton/coulomb or volt/metre. Example What is the electric field strength between a pair of parallel conducting plates, 2 mm apart, when the potential difference between them is 200 V? E = V/d = 200/(2 x 10 −3) = 1.0 x 10 5 V/m. 13.2.1 Capacitance Figure 13.3 Charging a capacitor When a pair of parallel conducting plates are connected to a d.c. supply and a potential difference produced between them, one of the plates becomes positively charged and the other negatively charged (Figure 13.3)...
eBook - ePub- C R Robertson(Author)
- 2008(Publication Date)
- Routledge(Publisher)
...3.4. Fig. 3.4 Note that the electric field will exist in all of the space surrounding the two plates, but the uniform section exists only in the space between them. Some non-uniformity is shown by the curved lines at the edges (fringing effect). At this stage we are concerned only with the uniform field between the plates. If a positively charged particle was placed between the plates it would experience a force that would cause it to move from the positive to the negative plate. The value of force acting on the particle depends upon what is known as the electric field strength. 3.3 Electric Field Strength (E) This is defined as the force per unit charge exerted on a test charge placed inside the electric field. (An outdated name for this property is ‘electric force’). 3.4 Electric Flux (ψ) and Flux Density (D) In the SI system one ‘line’ of flux is assumed to radiate from the surface of a positive charge of one coulomb and terminate at the surface of a negative charge of one coulomb. Hence the electric flux has the same numerical value as the charge that produces it. Therefore the coulomb is used as the unit of electric flux. In addition, the Greek letter psi is usually replaced by the symbol for charge, namely Q. The electric flux density D is defined as the amount of flux per square metre of the electric field. This area is measured at right angles to the lines of force. This gives the following equation Worked Example 3.1 Q Two parallel plates of dimensions 30 mm by 20 mm are oppositely charged to a value of 50 mC. Calculate the density of the electric field existing between them. A Q = 50 × 10 −3 C; A = 30 × 20 × 10 −6 m 2 Worked Example 3.2 Q Two parallel metal plates, each having a csa of 400 mm 2, are charged from a constant current source of 50 μA for a time of 3 seconds...
eBook - ePubElectrical Engineering
Fundamentals
- Viktor Hacker, Christof Sumereder(Authors)
- 2020(Publication Date)
- De Gruyter Oldenbourg(Publisher)
...The sources of an electric field are positive charges, sinks and negative charges. The electric field [Vm −1 ; NC −1 ] is uniform when neither its magnitude nor its direction changes from one point to another. 1.7.1 Force on charged particle in electric field The electric field strength is the measure of the force, a charged body experiences in an electric field. The existing electric field strength at any point in the field can be defined as the force exerted on a positive point unit charge Q + = 1 A s located there. Defining equation: F ⃗ = E ⋅ q The force causes an acceleration of the charged particle with the mass m a ⃗ = F ⃗ m 1.7.2 Force between point charges Coulomb’s law describes the force F between two point charges or spherically distributed charges Q 1 and Q 2 within a medium. The attracting or repelling forces F of individual charges are very small. They were measured for the first time by the French physicist Coulomb using a highly sensitive torsion balance. The spherical distribution of field lines around charges results in Coulomb’s law : F = 1 4 ⋅ π ⋅ ε ⏟ p r o p o r t i o n a l i t y − c o n s t a n t ⋅ Q 1 ⋅ O 2 r 2 with ε = ε 0 ⋅ ε r F Force in N Q 1, Q 2 Charge in As ε 0 Absolute dielectric constant of the vacuum ε 0 = 8.859 ⋅ 10 − 12 F m ε r Relative dielectric. constant 13 (indicates the factor of deviation of the dielectricity from the vacuum) for air is ε r = 1.00059 r Distance between the charges in m Table 1.4: Examples of ε r. Medium ε r Vacuum 1 (by definition) Air 1.00059 Org. insulators 2 to 4 Glasses 5 to 10 Water 81 1.8 Electric potential φ Electric potential φ is a scalar quantity equal to the amount of work needed to move a unit charge from a reference point (e.g. earth) to a specific point inside the electric field. In Figure 1.15 two horizontally charged plates and one positively charged object on the lower plate are shown. The potential energy of the object on the lower level 1 is W p, 1 and on the upper level 2 W p, 2...
eBook - ePubElectromagnetics Explained
A Handbook for Wireless/ RF, EMC, and High-Speed Electronics
- Ron Schmitt(Author)
- 2002(Publication Date)
- Newnes(Publisher)
...2 FUNDAMENTALS OF ELECTRIC FIELDS THE ELECTRIC FORCE FIELD To understand high-frequency and RF electronics, you must first have a good grasp of the fundamentals of electromagnetic fields. This chapter discusses the electric field and is the starting place for understanding electromagnetics. Electric fields are created by charges; that is, charges are the source of electric fields. Charges come in two types, positive (+) and negative (–). Like charges repel each other and opposites attract. In other words, charges produce a force that either pushes or pulls other charges away. Neutral objects are not affected. The force between two charges is proportional to the product of the two charges, and is called Coulomb’s law. Notice that the charges produce a force on each other without actually being in physical contact. It is a force that acts at a distance. To represent this “force at distance” that is created by charges, the concept of a force field is used. Figure 2.1 shows the electrical force fields that surround positive and negative charges. Figure 2.1 Field lines surrounding a negative and a positive charge. Dotted lines show lines of equal voltage. By convention, the electric field is always drawn from positive to negative. It follows that the force lines emanate from a positive charge and converge to a negative charge. Furthermore, the electric field is a normalized force, a force per charge. The normalization allows the field values to be specified independent of a second charge. In other words, the value of an electric field at any point in space specifies the force that would be felt if a unit of charge were to be placed there. (A unit charge has a value of 1 in the chosen system of units.) Electric field = Force field as “felt” by a unit charge To calculate the force felt by a charge with value, q, we just multiply the electric field by the charge, The magnitude of the electric field decreases as you move away from a charge, and increases as you get closer...
eBook - ePubRadiation Detection
Concepts, Methods, and Devices
- Douglas McGregor, J. Kenneth Shultis(Authors)
- 2020(Publication Date)
- CRC Press(Publisher)
...However, it is more common to use the general term potential, which is defined as the potential energy per unit charge, V = U Q ′ = 1 4 π ϵ o ∑ i = 1 n Q i r i, (8.16) and is expressed in units of volts (or joules per coulomb). Note that the potential is no longer dependent upon the “test” charge Q ′. 3 The force exerted upon Q ′ may also be expressed in terms of the electric field, produced by one or more point charges, in which F = Q ′ E. Substitution of q ′ E into Eq. (8.13) and division by Q ′ gives the potential difference between two points within the electric field. Hence, the potential difference between arbitrary locations a and b is V a b = Δ V = ∫ a b E ⋅ d l = ∫ a b E cos θ d l. (8.17) In summary, Eq. (8.17) is the voltage that an experimenter would measure between two points (a and b) within an electric field. The work done on a unit test charge moving from some point a to another point b in the electric field is Q ′ V ab. 8.3 Capacitance Consider the arrangement depicted in Fig. 8.6. Two conductive plates, separated by a distance d, have equal, but opposite, charges. An electric field is produced between the plates by the charges on the plates. The positively charged plate (or terminal) has a voltage V 1 and the negatively charged plate has a voltage V 2. The capacitance of the two plates is defined as the ratio of the charge magnitude on either plate to the magnitude of the potential difference between the plates, C = | Q Δ V |. (8.18) If Δ V is taken as the applied voltage V between the electrodes, then the above definition gives the important relation C V = Q (8.19) The SI unit for capacitance is the farad (one coulomb per volt)...
- G. Jagadeeswar Reddy, T. Jayachandra Prasad(Authors)
- 2020(Publication Date)
- CRC Press(Publisher)
...Chapter 1 Static Electric Fields 1.1 Introduction Electrostatic in the sense static or rest or time in-varying electric fields. Electrostatic field can be obtained by the distribution of static charges. The two fundamental laws which describe electrostatic fields are Coulomb’s law and Gauss’s law: They are independent laws. i.e., one law does not depend on the other law. Coulomb’s law can be used to find electric field when the charge distribution is of any type, but it is easy to use Gauss’s law to find electric field when the charge distribution is symmetrical. 1.2 Coulomb’s Law This law is formulated in the year 1785 by Coulomb. It deals with the force a point charge exerts on another point charge; generally a charge can be expressed in terms of coulombs. 1 coulomb = 6 × 10 18 electrons 1 electron charge = − 1.6 × 10 − 19 Coulombs Coulomb’s law states that the force between two point charges Q 1 and Q 2 is along the line joining between them, directly proportional to the product of two point charges, and inversely proportional to the square of the distance between them ∴ F = K Q 1 Q 2 R 2 where K is proportional constant In SI, a unit for Q 1 and Q 2 is coulombs(C), for R meters(m) and for F. newtons(N). K = 1 4 π ∊ 0 where ∊ 0 = permittivity of free space (or) vacuum = 8.854 × 10 − 12 farads / meter = 10 − 9 36 π farads / m K = 36 π 4 π × 10 − 9 = 9 × 10 9 m / farads F = Q 1 Q 2 4 π ∊ 0 R 2 (1.2.1) Assume that the point charges Q 1 and Q 2 are located at (x 1, y 1,. z 1) and (x 2, y 2, z 2) with the position vectors r ¯ 1 and r ¯ 2 respectively. Let the force on Q 2 due to Q 1 be F ¯ 12 which can be written as F ¯ 12 = Q 1 Q 2 4 π ∊ 0 R 2 a ¯ R 12 (1.2.2) where a ¯ R 12 is unit vector along the vector R ¯ 12. Graphical representation of the vectors in rectangular coordinate system is shown in Fig.1.1 Fig...
