Electromagnetic Field Theory (EMT) Lecture # 9 1) Coulomb s Law and Field Intensity 2) Electric Fields Due to Continuous Charge Distributions Line Charge Surface Charge Volume Charge
Coulomb's Law Coulomb's law states that the force F between two point charges (Q 1 and Q 2 is: 1) Along the line joining them 2) Directly proportional to the product Q 1 Q 2 of the charges 3) Inversely proportional to the square of the distance R between them. So F = kq 1Q 2 R 2 4) In SI units, charges Q 1 and Q2 are in coulombs C, the distance R is in meters m, and the force F is in newtons (N) 5) k = 1/4πε 0. The constant ε 0 is known as the permittivity of free space (in farads per meter) and has the values: ε 0 = 8.85 10-12, k = 9 10-9 m/f
Coulomb's Law Permittivity: Permittivity is a measure of how an electric field affects, and is affected by, a dielectric medium. Permittivity relates to a material's ability to resist an electric field. Relative Permittivity: r 0 0 8.85 10 The relative permittivity of a material is its dielectric permittivity expressed as a ratio relative to the permittivity of vacuum Relative permittivity is the factor by which the electric field between the 12 n c v r charges is decreased relative to vacuum. r 0
Coulomb's Law
Coulomb's Law Hence, Coulomb's law after inserting k = 1/4πε 0 will be F = Q 1Q 2 4πε 0 R 2 If point charges Q 1 and Q 2 are located at points having position vectors r 1 and r 2, then the force F 12 on Q 2 due to Q 1, shown in Figure below. F 12 = Q 1Q 2 4πε 0 R 2 a R 12 R 12 = r 2 r 1, R = R 12, a R12 = R 12 R F 12 = Q 1Q 2 4πε 0 R 2 R 12 R = Q 1Q 2 4πε 0 R 3 R 12 F 12 = Q 1Q 2 r 2 r 1 4πε 0 r 2 r 1 3
Coulomb's Law The force F 12, on Q 1 due to Q 2 is given by F 12 = F 21 The distance R between the charged bodies Q 1 and Q 2 must be large compared with the linear dimensions of the bodies; i.e., Q 1 due to Q 2 must be point charges. Q 1 due to Q 2 must be static (at rest). The signs of Q 1 due to Q 2 must be taken into account.
Principle of Superposition: Coulomb's Law For N charges Q 1, Q 2,. Q N located at points with position vectors r 1, r 2,. r N, the resultant force F on the charge Q located at point r is the vector sum of the forces exerted on Q by each of the charges Q 1, Q 2,. Q N. F = QQ 1 r r 1 4πε 0 r r 1 3 + QQ 1 r r 2 4πε 0 r r 2 3 + + QQ 1 r r N 4πε 0 r r N 3 F = Q 4πε 0 N k=1 Q k r r k r r k 3
Electric Field Intensity The electric field intensity (or electric field strength) E is the force per unit charge when placed in the electric field. E = F Q = Q 4πε 0 R 2 a R = Q r r 4πε 0 r r 3 4) The electric field intensity E is obviously in the direction of the force F and is measured in newtons/coulomb or volts/meter. 5) For N point charges Q 1, Q 2,. Q N located at r 1, r 2,. r N, the electric field intensity at point r is obtained. F = Q 1 r r 1 4πε 0 r r 1 3 + Q 1 r r 2 4πε 0 r r 2 3 + + Q 1 r r N 4πε 0 r r 3 = 1 N 4πε 0 N k=1 Q k r r k r r k 3
Electromagnetic Field Theory (EMT) Lecture # 9 1) Coulomb s Law and Field Intensity 2) Electric Fields Due to Continuous Charge Distributions Line Charge Surface Charge Volume Charge
Introduction So far we have only considered forces and electric fields due to point charges, which are essentially charges occupying very small physical space. It is also possible to have continuous charge distribution along a line, on a surface, or in a volume, as shown below: E = L ρ L dl 4πε 0 R 2 a R E = Q 4πε 0 R 2 a R E = S ρ s ds 4πε 0 R 2 a R v ρ v dv 4πε 0 R 2 a R
Electric Fields Due to Line Charge It is customary to denote the line charge density, surface charge density, and volume charge density by ρ L (in C/m), ρ s (in C/m 2 ), and ρ v (in C/m 3 ), respectively These must not be confused with ρ (without subscript) used for radial distance in cylindrical coordinates The charge element dq and the total charge Q due to line charge distribution is obtained as:
Electric Fields Due to Line Charge Practical example of a line charge distribution is a charged conductor of very small radius and a sharp beam in a cathode-ray tube.
Electric Fields Due to Line Charge In the case of the electron beam the charges are in motion and it is true that we do not have an electrostatic problem However, 1. If the electron motion is steady and uniform (a DC beam) and 2. If we ignore for the moment the magnetic field which is produced. The electron beam may be considered as composed of stationary electrons.
Line Charge Distribution The equation for electric field due to point charge is: The electric field intensity due to line charge distribution ρ L may be regarded as the summation of the field contributed by the numerous point charges making up the charge distribution Thus by replacing Q in the equation with charge element dq = ρ L dl, we get: We shall now apply this formula to line charge distribution
Line Charge Distribution Consider a line charge with uniform charge density ρ L extending from A to B along the z-axis as shown in figure below:
Line Charge Distribution Thus for a finite line charge, we have: (Derivation given in the book) As a special case, for an infinite line charge, point B is at (0,0, ) and A at (0,0, ) So 1 = π 2, 2= π 2 and the z-component vanishes (How?) The above equation reduces to the equation below: Note: ρ is the perpendicular distance from the line to the point of interest
Problem-1 A circular ring of radius a carries a uniform charge ρ L C/m and is placed on the xy-plane with axis the same as the z-axis. Calculate the electric field intensity E on z-axis at a distance of h from the origin.
Electromagnetic Field Theory (EMT) Lecture # 9 1) Coulomb s Law and Field Intensity 2) Electric Fields Due to Continuous Charge Distributions Line Charge Surface Charge Volume Charge
Introduction Another basic charge configuration is the infinite sheet of charge having a uniform density of ρ s C/m 2 Such a charge distribution may often be used to approximate the charge found on the conductors of a strip transmission line or a parallel-plate capacitor For ease of derivation, the sheet of charge is considered to be infinite This is a good approximation since the distances involved in the measurement of fields are generally small compared to the dimensions of the sheet of charge
Consider an infinite sheet of charge in the xy-plane with uniform charge density ρ s. Sheet of Charge The charge associated with an elemental area ds is: dq = ρ s ds where, ds = dydz and hence the total charge is Q = ρ s ds
From E = S ρ s ds a 4πε 0 R 2 R Sheet of Charge, The contribution to the electric filed E at point (0, 0, h) by the charge dq on the element surface 1 shown in Figure below is. de = dq 4πε o R 2 a R Where, R = ρ a ρ + ha z, R = R = ρ 2 + h 2 1 2 a R = R R, dq = ρ sds = ρ s ρ dφ dρ
Sheet of Charge Substitutin R = R = ρ 2 + h 2 1 2, a R = R R, dq = ρ sds = ρ s ρ dφ dρ in de = dq 4πε o R 2 a R we ll get following de = ρ s ρ dφ dρ ρ a ρ + h a z 4πε o ρ 2 + h 2 3 2 Owing to symmetry of the charge distribution, for every element 1, there is a corresponding element 2 whose contribution along a ρ cancels that of element 1 as shown in the Figure. Thus E has only z component.
Sheet of Charge Substitutin R = R = ρ 2 + h 2 1 2, a R = R R, dq = ρ sds = ρ s ρ dφ dρ in de = dq 4πε o R 2 a R we ll get following de = ρ s ρ dφ dρ ρ a ρ + h a z 4πε o ρ 2 + h 2 3 2 Owing to symmetry of the charge distribution, for every element 1, there is a corresponding element 2 whose contribution along a ρ cancels that of element 1 as shown in the Figure. Thus E has only z component.
Sheet of Charge There fore, E = S de z = ρ s 4πε o 2π φ=0 ρ=0 h ρ dφ dρ ρ 2 + h 2 3 2 a z E = ρ s 2ε o a z This Eq. is valid for h > 0 and tells that E has only z-component if the charge is in xy-plane. For h < 0, we need to replace a z with a z. In general for an infinite sheet of charge E = ρ s 2ε o a n Where a n is unit vector normal to the sheet.
A circular disk of radius a is uniformly charged with ρ s C/m 2. If the disk lies on the z = 0 plane with its axis along the z-axis, (a) Show that at point (0, 0, h) Problem-1 (b) From this, derive the E field due to an infinite sheet of charge on the z = 0 plane
Electromagnetic Field Theory (EMT) Lecture # 9 1) Coulomb s Law and Field Intensity 2) Electric Fields Due to Continuous Charge Distributions Line Charge Surface Charge Volume Charge
Volume Charge A volume charge may be visualized by a region of space with a large number of charges separated by very small distances We can replace this distribution of very small particles with a smooth continuous distribution described by a volume charge density We denote the volume charge density by ρ v, having the units of coulombs per cubic meter (C/m 3 )
Volume Charge A volume charge distribution with uniform charge density ρ v is shown in figure We choose a volume of the shape of a cube
Volume Charge The charge dq associated with the elemental volume dv is The total charge is given as: Here, elemental volume dv depends upon the shape of the volume charge
Volume Charge From figure, the electric field de outside the sphere at P(0,0,z) due to the elementary volume dv is: de = ρ vdv 4πε o R 2 a R
Volume Charge From the figure, a R may be written as: Due to the symmetry of the charge distribution, the contributions to E x or E y add up to zero We are left with only E z, given by: We need to derive expressions for dv, R 2, and cosα
Volume Charge Applying the cosine rule (given in the figure below) to the figure, we have: It is convenient to evaluate the integral in terms of R and r Hence we express cos θ', cos α, and sin θ' dθ' in terms of R and r', that is:
Volume Charge Differentiating the above equation with respect to θ' keeping z and r' fixed, we obtain: Substituting values in the integral, we get:
Volume Charge Earlier, the total charge was given as: If we assume that the whole volume charge has a spherical volume with radius a, then: Using this value in the equation for electric field, we get:
Volume Charge Due to the symmetry of the charge distribution, the electric field at P(r,θ,Φ) is readily obtained as: It may be observed that the above equation is identical to the electric field at the same point due to a point charge Q located at the origin or the center of the spherical charge distribution.
Problem-1 Find the total charge contained in a 2 cm length of the electron beam shown in the figure.
Problem-2 Calculate the total charge within each of the indicated volume: a) 0 ρ 0.1, 0 π, 2 z 4; ρ v = ρ 2 z 2 sin0.6 b) Universe; ρ v = e 2r r 2