Coulomb's and Gauss's Laws - Electromagnetics for Medical Physics - Al-Farahidi University - PDF

Coulomb's and Gauss' laws Electromagnetics for medical physics Second year ‫الكهرومغناطيسية للفيزياء الطبية المستوى الثاني‬ The third lecture ‫المحاضرة الثالثة‬ Dr. Faten Monjed Hussein 1 Point Charge ⚫ The term point charge refers to a particle...

Coulomb's and Gauss' laws Electromagnetics for medical physics Second year ‫الكهرومغناطيسية للفيزياء الطبية المستوى الثاني‬ The third lecture ‫المحاضرة الثالثة‬ Dr. Faten Monjed Hussein 1 Point Charge ⚫ The term point charge refers to a particle of zero size that carries an electric charge ⚫ The electrical behavior of electrons and protons is well described by modeling them as point charges ⚫ The electrical force between two stationary point charges is given by Coulomb’s Law ⚫ The force is inversely proportional to the square of the separation r between the charges and directed along the line joining them ⚫ The force is proportional to the product of the charges, q1 and q2, on the two particles ⚫ The force is attractive if the charges are of opposite sign ⚫ The force is repulsive if the charges are of like sign ⚫ The force is a conservative force ⚫ Mathematically: |𝑞1 ||𝑞2 | 𝐹 = 𝑘𝑒 𝑟2 The SI unit of charge is the coulomb (C) ke is called the Coulomb constant ke = 8.9876 x 109 N.m2/C2 = 1/(4π𝜀 o) 𝜀 o is the permittivity of free space 𝜀 o = 8.8542 x 10-12 C2 / N.m2 ⚫ The like charges produce a repulsive force between them ⚫ Electrical forces obey Newton’s Third Law ⚫ ⃗⃗⃗⃗⃗⃗ 𝐹12 = −𝐹 ⃗⃗⃗⃗⃗⃗ 21 ⚫ With unlike signs for the charges, the product q1q2 is negative and the force is attractive 2 EX The figure below shows situation in which charged particles are at equal distances (r) from the origin. Assume r = 2 cm and q 3 = −2 nC , q 2 = +10 nC , q1 = +8 nC. Find the net electric force exerted on𝒒𝟏. Solution |𝒒𝟏 ||𝒒𝟐 | 𝐹21 = 𝑲𝒆 (𝟐𝒓𝟐 ) |𝟖 × 𝟏𝟎−𝟗 ||𝟏𝟎 × 𝟏𝟎−𝟗 | 𝑭𝟐𝟏 = 𝟗 × 𝟏𝟎𝟗 × (𝟐(𝟐 × 𝟏𝟎−𝟐 )𝟐 ) = 90 × 10−5 𝑁 𝟗 |𝟖 × 𝟏𝟎−𝟗 ||−𝟐 × 𝟏𝟎−𝟗 | 𝑭𝟑𝟏 = 𝟗 × 𝟏𝟎 × (𝟒 × 𝟏𝟎−𝟐 )𝟐 = 9 × 10−5 𝑁 𝜽𝟐𝟏 = 𝟏𝟖𝟎 − 𝟒𝟓 = 𝟏𝟑𝟓° , 𝜽𝟑𝟏 = 𝟎° 𝐹𝑛𝑒𝑡 = (90𝑐𝑜𝑠0 + 9𝑐𝑜𝑠135) × 10−5 𝑖̂ + (90𝑠𝑖𝑛0 + 9𝑠𝑖𝑛135) × 10−5 𝑗̂ 𝐹𝑛𝑒𝑡 = (39.09) × 10−5 𝑖̂ + (6.364) × 10−5 𝑗̂ 𝟐 𝑭𝒕𝒐𝒕 = √(𝟑𝟗. 𝟎𝟗)𝟐 + (𝟔. 𝟑𝟔𝟒)𝟐 × 𝟏𝟎−𝟓 = 𝟑𝟗. 𝟔𝟖 × 𝟏𝟎−𝟓 𝑵 3  Gauss’s Law Introduction: ⚫ Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the field 𝜙 𝐸 = 𝐸⃗. 𝐴 ⚫ The field lines may make some angle θ with the perpendicular to the surface ⚫ 𝜙𝐸 = 𝐸𝐴𝑐𝑜𝑠𝜃 ⚫ The units of electric flux will be N.m2/C ⚫ Gauss’s law is an expression of the general relationship between the net electric flux through a closed surface and the charge enclosed by the surface ⚫ The closed surface is often called a Gaussian surface ⚫ Gauss’s law is of fundamental importance in the study of electric fields 𝒒 ⃗. ⃗⃗⃗⃗⃗ ⚫ 𝚽𝒆 = ∮ ⃗𝑬 𝒅𝑨 = 𝑬𝒏 ∮ 𝒅𝑨 = 𝒊𝒏 𝜺𝒐 - qin is the net charge inside the surface, 4  - 𝑬𝒏 the normal component of ⃗⃗⃗𝑬 on surface vector ⚫ ⃗⃗⃗𝑬 represents the electric field at any point on the surface ⚫ Although Gauss’s law can, in theory, be solved to find ⃗⃗⃗𝑬 for any charge configuration, in practice it is limited to symmetric situations. ⚫ the net flux through any closed surface surrounding a point 𝐪 charge q is given by 𝐢𝐧 and is independent of the shape of the 𝛆𝟎 surface ⚫ the electric flux through a closed surface that surrounds no charge is zero EX.1 Find electric field due to spherically symmetric charge distribution (Q) (charge volume density is uniform), on insulator sphere which has radius (a), at point p where ;1) r>a, and at 2) r