Lecture 2 PDF - Electric Fields

For Faculty of Applied Health Science Technology Faculty of Computer Science and Engineering By Dr. Mohamed EL-Henawey Electric Fields Electric Field: An electric field is said to exist in the region of space around a charged object. This charged objec...

For Faculty of Applied Health Science Technology Faculty of Computer Science and Engineering By Dr. Mohamed EL-Henawey Electric Fields Electric Field: An electric field is said to exist in the region of space around a charged object. This charged object is the source charge. The presence of the electric field can be detected by placing a test charge in the field and noting the electric force on it. As an example, consider a small positive test charge 𝒒𝟎 placed near a second object carrying a much greater positive charge Q. Electric Field: We define the electric field due to the source charge at the location of the test charge to be the electric force on the test charge per unit charge, or, to be more specific, the electric field vector 𝑬 is defined as the electric force 𝑭𝒆 acting on a positive test charge 𝒒𝟎 divided by the test charge. 𝑭𝒆 𝑵 𝑬= 𝒒𝟎 𝑪 The direction of 𝑬 is that of the force on a positive test charge. 𝑭𝒆 = 𝒒𝟎 𝑬 The Electric Field of a Point Charge: According to Coulomb’s law, 𝒒 𝒒𝟎 𝑭𝒆 = 𝒌 𝟐 𝒓 𝑬= 𝑭𝒆 = 𝒒 𝒌 𝟐 𝒒 𝑵 𝒒𝟎 𝒓 𝑬= 𝒌 𝟐 𝒓 𝑪 Electric Field Direction: q is positive, the force is directed away from q. The direction of the field is also away from the positive source charge. q is negative, the force is directed toward q. The field is also toward the negative source charge. Electric Field Direction: Electric Fields from Multiple Charges: At any point P, the total electric field due to a group of source charges equals the vector sum of the electric fields of all the charges. ՜ 𝒒𝒊 ∧ 𝑬 = 𝑬𝟏 + 𝑬𝟐 + 𝑬𝟑 +....... = 𝑲 ෍ 𝟐 𝒓𝒊 𝒓𝒊 𝒊 Where is a unit vector directed from qi towards point P. ∧ 𝒓𝒊 Example (1): Find the electric force on a proton placed in an electric field of 2 × 104 N/C directed along the positive X-axis? Solution E =2 × 104 N/C & The charge of proton +e = 1.6 × 10−19 coulomb & F=???? Electric Fields : 𝑭 = 𝒒𝟎 𝐄 = 𝐞𝐄 = 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝐱 𝟐 × 𝟏𝟎𝟒 = 𝟑. 𝟐 × 𝟏𝟎−𝟏𝟓 𝐍 Example (2): A charge 𝒒𝟏 = 𝟕𝝁𝑪 is located at the origin. And a second charge 𝒒𝟐 = −𝟓𝝁𝑪 is located on the X-axis at 0.3m from the origin. Find the electric field at the point P with the coordinates (0, 0.4) m. Solution ‫شحنتين إحداهما موجبة و األخرى سالبة موضوعتين كما موضح بالشكل مطلوب حساب‬ P ‫المجال الكهربى عند النقطة‬ Electric Fields : A) Electric field for q1 is −6 q1 9 7 x10 E1 = K 2 = (9 x10 ) 2 = 3.94 x10 5 N /c r1 (0.4) And take the direction of positive Y-axis. B) Electric field for q2 is q2 5 x10 −6 E 2 = K 2 = (9 x10 ) 9 2 = 1.8 x10 5 N /c r2 (0.5) So, y-component of total E is: 0.4 E y = E1 − E 2 sin = 3.94 x10 5 − 1.8 x10 5 x = 2.5 x10 5 N / C 0.5 And x-component of total E is: 0.3 E x = E 2 cos = 1.8 x10 5 x = 1.1 x10 5 N / C 0.5 Electric Fields : Then, the total field E has the magnitude: E = E x + E y = (1.1 x105 ) 2 + (2.5 x105 ) 2 = 2.7 x105 N / C 2 2 And it's direction: Ey 2.5 x10 5 tan  = = = 2.27   = 66 o E x 1.1 x10 5 From this result, we find that E has a magnitude of 2.7 x 10 5 N/C and makes an angle  of 66° with the positive x axis. Electric Field – Continuous Charge Distribution: Procedure: 1. Divide the charge distribution into small elements, each of which contains Δq. 2. Calculate the electric field due to one of these elements at point P. 3. Evaluate the total field by summing the contributions of all the charge elements. For the individual charge elements 𝜟𝒒 Δ𝑬 = 𝑲 𝟐 𝒓ො 𝒓 Electric Field – Continuous Charge Distribution: Because the charge distribution is continuous 𝜟𝒒𝒊 𝒅𝒒 𝑬 = 𝑲෍ 𝟐 ො 𝒓 𝒊 = 𝑲 න 𝟐 𝒓ො 𝒓𝒊 𝒓 𝒊 Different cases of charge distribution If a charge Q is uniformly distributed throughout a volume V, the volume charge density ρ is defined by 𝑸 𝝆= (𝑪/𝒎𝟑 ) 𝑽 Charge Densities: If a charge Q is uniformly distributed on a surface of area A, the surface charge density σ is defined by 𝑸 𝝈= (𝑪/𝒎𝟐 ) 𝑨 If a charge Q is uniformly distributed along a line of length L, the linear charge density λ is defined by 𝑸 𝝀= (𝑪/𝒎) 𝒍 Charge Densities: Electric Field Lines: The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region. Therefore, the field lines are close together where the electric field is strong and far apart where the field is weak. The density of lines through surface A is greater than through surface B. The magnitude of the electric field is greater on surface A than B. Electric Field Lines: For a positive source charge, the lines will radiate outward. For a negative source charge, the lines will radiate inward. Electric Field Lines: Electric Field Lines – Dipole The charges are equal and opposite. The number of field lines leaving the positive charge equals the number of lines terminating on the negative charge. Electric Field Lines – Like Charges The charges are equal and positive. The same number of lines leave each charge since they are equal in magnitude. At a great distance, the field is approximately equal to that of a single charge of 2q. Since there are no negative charges available, the field lines end infinitely far away. Electric Field Lines: Electric Field Lines, Unequal Charges The positive charge is twice the magnitude of the negative charge. Two lines leave the positive charge for each line that terminates on the negative charge. At a great distance, the field would be approximately the same as that due to a single charge of +q. Motion of Charged Particles: If the particle has a positive charge, its acceleration is in the direction of the field. If the particle has a negative charge, its acceleration is in the direction opposite the electric field. Motion of an electric dipole in a uniform electric field : The electric dipole is placed in a uniform electric field E and its dipole moment P making an angle with E. Two equal and opposite forces act on the dipole (F) and (-F) as shown in the fig : Net force on the dipole = Zero The forces on the charge +q and - q resulting in a torque on the dipole Which rotates the dipole. Thus ,the torque tend to align the dipole axis along the direction of the field E. Torque T = Force (qE) x the perpendicular distance between the antiparallel forces Motion of an electric dipole in a uniform electric field :  =F xd but d = 2a sin    = 2 a q E sin = P E sin Where P = 2a q is the electric dipole moment