Chapter 1 Gauss' Law PDF

Chapter 1 GAUSS’ LAW  Contents Some physical concepts Electric Flux (φ) Formula of Gauss’ law Chapter 1 Coulombs from Gauss Electric field of cylindrical sy...

Chapter 1 GAUSS’ LAW  Contents Some physical concepts Electric Flux (φ) Formula of Gauss’ law Chapter 1 Coulombs from Gauss Electric field of cylindrical symmetry Electric field of charged planer symmetry Electric field due to charged sphere (Spherical symmetry)  Some physical concepts Gaussian surface; Gaussian surface; It is more symmetrical imaginary closed surface and surrounded the charge from all direction. It is suggested according the shape of electric charge distribution.  Closed integration ; ‫𝑨𝒅 ׯ‬ ❑ It is one kind of integration and apply on closed surface ❑ its value equal the area of Gaussian surface. ර 𝒅𝑨 = 𝟒𝝅𝒓𝟐 𝒔𝒑𝒉𝒆𝒓𝒆 ර 𝒅𝑨 = 𝟐𝝅𝒓 𝒉 𝒔𝒊𝒅𝒆 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒄𝒚𝒍𝒊𝒏𝒅𝒆𝒓 ර 𝒅𝑨 = 𝝅 𝒓𝟐 𝒄𝒊𝒓𝒄𝒖𝒍𝒆  Vector area ; it is vector its value is A (area) its direction outward and normal on Gaussian surface Scalar Product 𝑨 ⊙ 𝑩 = 𝑨 × 𝑩 𝒄𝒐𝒔𝜽 𝒊𝒇 𝜽 = 𝟎 ∴ 𝑨 ⊙ 𝑩 = 𝑨. 𝑩 𝒄𝒐𝒔𝟎 = 𝑨𝑩 𝒊𝒇 𝜽 = 𝝅 ∴ 𝑨 ⊙ 𝑩 = 𝑨. 𝑩 𝒄𝒐𝒔 𝝅𝒐 = −𝑨𝑩 𝒊𝒇 𝜽 = 𝟗𝟎𝒐 ∴ 𝑨 ⊙ 𝑩 = 𝑨. 𝑩 𝒄𝒐𝒔 𝟗𝟎𝒐 = 𝟎  What is Electric flux [𝚽] ? the total number of electric filed lines passing through the normal surface area placed at this point The electric flux is direct proportional with the magnitude of charge 𝑸𝒊𝒏 𝚽∝𝑸 ∴ 𝚽= 𝜺𝒐 Unit of electric flux is (N.m2/C) 𝚽 = 𝑬. 𝑨 = 𝑬 𝑨 𝒄𝒐𝒔 𝜽 Where: A is a vector normal to the surface (magnitude A, and direction normal to the surface).  is the angle between E and A For which of these closed surfaces (a, b, c, d) the flux of the electric field, produced by the charge +2q, is zero?  Calculate the flux of the electric field  for each of the closed surfaces a, b, c, and d Surface a, a = Surface b, b = Surface c, c = Surface d, d =  Gauss’ Law 𝑸𝒊𝒏 𝜱= = ‫𝑬 = 𝜽 𝒔𝒐𝒄 𝑨𝒅 𝑬 ׯ = 𝑨𝒅 ∙ 𝑬 ׯ‬. 𝑨𝒏𝒐𝒓𝒎𝒂𝒍 𝜺𝟎 𝚽 is electric flux, 𝑸𝒊𝒏 is sum of electric charge within gaussian surface, 𝜺𝟎 = 𝟖. 𝟖𝟓 × 𝟏𝟎−𝟏𝟐 𝒄𝟐 /𝐍𝒎𝟐 is electric permittivity Example (1) A point charge of 1.8 µc is at the center of a cubical Gaussian surface 55 cm on edge. What is the net electric flux through the surface? Solution 𝒒 𝟏. 𝟖𝒙𝟏𝟎−𝟔 𝟓 𝟐 𝜱= = −𝟏𝟐 = 𝟐 × 𝟏𝟎 𝑵. 𝒎 /𝑪 𝜺𝟎 𝟖. 𝟖𝟓𝟒𝒙𝟏𝟎 Calculate the flux of the electric field E, through the surface A, in each of the three cases shown: a)  =E A b)  =zero c) ∅ = 𝑬𝑨 𝐜𝐨𝐬 𝜽 Steps to derive or solve the Gauss law problem; ‫حل مسائل الشحنات الموزعة باستخدام قانون جاوس يتم االتى‬ 1. Detriment the shape of electric charge ‫تحديد شكل الشحنة‬ Point charge or charged sphere (Q) Charged line or charged cylinder 𝑸 = 𝝀𝒍 Charged surface (plate, plane, sheet) 𝑸 = 𝝈𝑨 2. Assume gaussian surface according to shape charge ‫إختيار سطح مغلق(جاوسى) الحتواء الشحنة الموزعة‬ 3. Apply Gauss law ‫تطبيق قانون جاوس‬ 𝑸𝒊𝒏 𝑮.𝑺 = ර 𝑬. 𝒅𝑨 = ර𝑬 𝒅𝑨 𝒄𝒐𝒔 𝜽 = 𝑬 ර 𝒅𝑨 = 𝑬(𝑨 𝑮.𝑺. ) 𝜺𝟎 4. Find the formula of electric field (E)  Deriving Coulomb’ s law from Gauss’ law To calculate the electric field at a distance (r) from a point charge q using Gauss’ law 1- Charge is a point charge (q) 2- Gaussian surface is a sphere of radius r 𝒒 = ර 𝑬. 𝒅𝑨 = ර𝑬 𝒅𝑨 𝒄𝒐𝒔 𝜽 = 𝑬 ර 𝒅𝑨 cos 𝜽 = 𝑬 𝑨𝑺𝒑𝒉𝒆𝒓𝒆 𝜺𝟎 𝒒 = 𝑬 𝟒𝝅𝒓𝟐 𝜺𝟎 𝟏 𝒒 E= 𝒒 𝟒𝝅𝜺𝒐 𝒓𝟐 ∴𝑬=𝒌 𝟐 𝒓 Derive the electric field at a distance (r) from a point charge (q) using Gauss’ law  Application of Gauss’s Law A Cylindrically Symmetry The electric field at a distance (r) from an infinite line of charges of constant charge per unit length λ 1- Charge is a liner charge (𝑸𝒊𝒏 = 𝝀𝒍) ) 2- Gaussian surface is a cylinder of radius r and Length (L) 𝑸𝒊𝒏 = ර 𝑬. 𝒅𝑨 = ර𝑬 𝒅𝑨 𝒄𝒐𝒔 𝜽 = 𝑬 ර 𝒅𝑨 = 𝑬𝑨 𝑨𝒏𝒐𝒓 = 𝟐𝝅𝒓𝒍 𝜺𝟎 𝝀𝒍 𝝀 𝝀 𝑬(𝟐𝝅𝒓𝒍) = 𝑬= = 𝟐𝒌 𝜺𝟎 𝟐𝝅𝜺𝟎 𝒓 𝒓 𝝀 𝑬 = 𝟐𝒌 𝒓 Derive the electric field at a distance (r) from a cylinder () using Gauss’ law Example (2) An infinite line of charge produces a field of 4.5 x 104 N/C at a distance of 2 m. Calculate the linear charge density. Solution 𝝀 𝑬= 𝟐𝝅𝜺𝟎 𝒓 𝝀 = 𝟐𝝅𝜺𝟎 𝒓𝑬 𝝀 = 𝟐𝝅 𝟖. 𝟖𝟓 × 𝟏𝟎−𝟏𝟐 𝟐 𝟒. 𝟓 × 𝟏𝟎𝟒 = 𝟓𝝁𝑪/𝒎 A planer Symmetry (conducting) To find the electric field due to an infinite conducting sheet has a uniform surface charge density σ The charge is a uniform surface charge density 𝑸𝒊𝒏 = 𝝈𝑨 the Gaussian surface is a cylinder one of its base on the plate. The electric field is zero everywhere inside the conductor 𝑸𝒊𝒏 = ර 𝑬. 𝒅𝑨 = ර𝑬 𝒅𝑨 𝒄𝒐𝒔 𝜽 = 𝑬 ර 𝒅𝑨 = 𝑬𝑨 𝜺𝟎 𝑸𝒊𝒏 =𝑬𝑨 𝜺𝟎 𝝈𝑨 ∴ =𝑬𝑨 𝜺𝟎 𝝈 𝑬= 𝜺𝟎 A planer Symmetry (non-conducting) To find the electric field due to an infinite non-conducting sheet has a uniform surface charge density σ The charge is a uniform surface charge density 𝑸𝒊𝒏 = 𝝈𝑨 σ Gaussian surface as in figure 𝑸𝒊𝒏 = 𝟐 ර 𝑬. 𝒅𝑨 = 𝟐 ර𝑬 𝒅𝑨 𝒄𝒐𝒔 𝜽 = 𝟐𝑬 ර 𝒅𝑨 = 𝑬𝑨 𝜺𝟎 𝝈𝑨 = 𝑬 𝟐𝑨 𝜺𝟎 𝝈 ∴𝑬= 𝟐𝜺𝟎  The E of two parallel non-conducting sheets 𝝈 𝑬= (i) between the sheets 𝟐𝜺𝟎 𝝈(−) 𝝈(+) 𝑬𝒃𝒆𝒕 = 𝑬(−) + 𝑬(+) = + 𝟐𝜺𝟎 𝟐𝜺𝟎 𝜎(+) 𝜎(−) 𝐸(+) + 𝐸(+) -𝐸 (+) (ii) to the left of the sheets + - + - + - 𝝈(−) 𝝈(+) - 𝑬𝒍𝒆𝒇𝒕 = 𝑬(−) − 𝑬(+) = − + 𝟐𝜺𝟎 𝟐𝜺𝟎 𝐸(−) + 𝐸(−) - 𝐸(−) + - + - (iii) to the right of the sheets 𝝈(+) 𝝈(−) 𝑬𝒓𝒊𝒈𝒉𝒕 = 𝑬(+) − 𝑬(−) = − 𝟐𝜺𝟎 𝟐𝜺𝟎 Non- uniform Electric field (E) The values of the electric field (E) depend on the distance (r) and /or the position of the point at which calculating the field. Such as ; (i) the electric field (E) of point or charged sphere, (ii) the electric field (E) of cylindrical symmetry 𝑸 𝑬=𝒌 𝟐 𝝀 𝒓 𝑬 = 𝟐𝒌 𝒓 Uniform Electric field (E) The values of the electric field (E) is constant and not depend on the distance (r) and /or the position of the point at which calculating the field. Such as ; Planner symmetry, the electric field (E) is constant, and depend on the value of surface charge density and kind of sheet (conductor or non-conductor). 𝝈 𝝈 ∴𝑬= ∴𝑬= 𝜺𝟎 𝟐𝜺𝟎 Example (3) The figure shows portions of two large non-conducting sheets, each with a fixed uniform charge on one side. The magnitude of the surface charge densities are += 6.8 C/m2 and -= 4.3 C/m2. Find the Electric field ; (i) to the left of the sheets, (ii) between the Sheets, and (III) to the right of the sheets 𝜎(+) 𝜎(−) (i) to the left of the sheets 𝝈(−) 𝝈(+) 𝟏𝟎−𝟔 𝟓 𝑬𝒍𝒆𝒇𝒕 = 𝑬(−) − 𝑬(+) = − = −𝟏𝟐 [𝟒. 𝟑 − 𝟔. 𝟖] = −𝟏. 𝟒 × 𝟏𝟎 𝑵/𝑪 𝟐𝜺𝟎 𝟐𝜺𝟎 𝟐(𝟖. 𝟖𝟓𝟒 × 𝟏𝟎 ) (ii) between the sheets 𝝈(−) 𝝈(+) 𝟏𝟎−𝟔 (𝟒. 𝟑 + 𝟔. 𝟖) 𝟓 𝑵/𝑪 𝑬𝒃𝒆𝒕 = 𝑬(−) + 𝑬(+) = + = = 𝟔. 𝟐 × 𝟏𝟎 Towards +ve X-axis 𝟐𝜺𝟎 𝟐𝜺𝟎 𝟐(𝟖. 𝟖𝟓𝟒 × 𝟏𝟎−𝟏𝟐 ) (iii) to the right of the sheets 𝝈(+) 𝝈(−) 𝟏𝟎−𝟔 (𝟔. 𝟖 − 𝟒. 𝟑) 𝟓 𝑵/𝑪 Towards +ve X-axis 𝑬𝒓𝒊𝒈𝒉𝒕 = 𝑬(+) − 𝑬(−) = − = = 𝟏. 𝟒 × 𝟏𝟎 𝟐𝜺𝟎 𝟐𝜺𝟎 𝟐(𝟖. 𝟖𝟓𝟒 × 𝟏𝟎−𝟏𝟐 ) Example (4) The magnitude of the average electric field normally present in the earth’ s atmosphere just above the surface of the earth is about 150 N/C, directed downward. What is the total net surface charge of the earth? Assume the earth to be a conductor with a uniform surface charge density. Assume 𝐫𝐚𝐝𝐢𝐮𝐬 𝐨𝐟 𝐞𝐚𝐫𝒕𝒉 𝒊𝒔 𝟔. 𝟑𝟕 × 𝟏𝟎𝟑 𝒌𝒎 solution 𝑬 = 𝝈Τ𝜺𝟎 𝝈 = 𝜺𝟎 × 𝑬 = 𝟖. 𝟖𝟓 × 𝟏𝟎−𝟏𝟐 𝟏𝟓𝟎 = 𝟏. 𝟑𝟑 × 𝟏𝟎−𝟗 𝑪/𝒎𝟐 𝒒 = 𝝈𝑨 = 𝝈 𝟒𝝅𝒓𝟐 = (𝟏. 𝟑𝟑 × 𝟏𝟎−𝟗 )(𝟒𝝅)(𝟔. 𝟑𝟕 × 𝟏𝟎𝟔 )𝟐 𝒒 = −𝟔. 𝟖 × 𝟏𝟎𝟓 𝑪 The negative sign indicate the earth has negative charge Example (5) In the figure, a small, non-conducting ball of mass m = 1 mg and uniformly distributed charge q = 2 x10-8 C hangs from an insulating thread that makes an angle 𝜽 = 𝟑𝟎𝒐 with a vertical, uniformly charged non-conducting sheet. Considering the ball’ s weight and assuming that the sheet extends far in all directions, calculate the surface charge density σ of the sheet. solution The sheet is non-conducting, so 𝝈 𝒒𝝈 𝑬= ⇒ ∴ 𝑭 = 𝑬𝒒 = 𝟐𝜺𝟎 𝟐𝜺𝟎 𝒒𝝈 ∴ ෍ 𝑭𝒙 = 𝟎 ⇒ 𝑻 𝐬𝐢𝐧 𝜽 = 𝟐𝜺𝟎 ∴ ෍ 𝑭𝒚 = 𝟎 ⇒ 𝑻 𝐜𝐨𝐬 𝜽 = 𝒎𝒈 Divided both equation, then 𝑻 𝒔𝒊𝒏 𝜽 𝒒𝝈 = 𝑻 𝒄𝒐𝒔 𝜽 𝟐𝒎𝒈𝜺𝟎 𝒒𝝈 ∴ 𝒕𝒂𝒏 𝜽 = ⇒ 𝟐𝒎𝒈𝜺𝟎 𝟐𝒎𝒈𝜺𝟎 𝒕𝒂𝒏 𝜽 ∴𝝈= 𝒒 𝟐 𝟏𝟎−𝟔 𝟗. 𝟖 𝟖. 𝟖𝟓 × 𝟏𝟎−𝟏𝟐 𝒕𝒂𝒏 𝟑𝟎 ∴𝝈= 𝟐 × 𝟏𝟎−𝟖 ∴ 𝝈 = 𝟓𝒏𝑪/𝒎𝟐 Metal (conductor) The metal material has free electrons (charges) that can move within the conductor and concentrating on its surface to attain the minimum energy Once the charges created in metal, they migrate to the surface of the metal, therefore the net charges inside the conductor are zero. So, the electric field (E) within the conductor is Zero The electric field is zero everywhere inside the conductor Non-Metal (Non-conductor) The non-metal material has no free electrons (charges), So the can not move within the material and formed localized charges within non-conductor The electric field (E) has values inside the non-conductor  Field (E) inside non conducting charged Sphere To find the electric field inside an insulating solid sphere of radius a which has a uniform volume charge density ρ and carries a total charge Q First inside the sphere The Gaussian surface is a sphere of radius r ˂ a ∵ 𝝆𝑪𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒏𝒈 𝑺𝒑𝒉𝒆𝒓𝒆 = 𝝆𝑮𝒐𝒖𝒔𝒔𝒊𝒂𝒏 𝒔𝒖𝒓𝒇𝒂𝒄𝒆 𝑸 𝒒 𝑸 𝟑 ∵𝝆= ∴ = 𝒓 𝑽 𝟒 𝟑 𝟒 𝟑 ∴ 𝒒 = 𝟑𝑸 𝝅𝒓 𝝅𝒂 𝒂 𝟑 𝟑 𝒒 ∴𝑬=𝒌 𝟐 The field due to small sphere is 𝒓 𝑸 𝒓𝟑 Substitute by q in above equation ∴𝑬=𝒌 𝟐 𝟑 𝒓 𝒂 𝑸 ∴ 𝑬 = 𝒌 𝟑𝒓 𝒂 E on the surface of the sphere At the surface of the sphere r = a 𝑸 𝑸 𝑬= =𝒌 𝟐 𝟒𝝅𝜺𝟎 𝒂𝟐 𝒂 Third outside the sphere the Gaussian surface is a sphere of radius r ˃ a 𝟏 𝑸 𝑸 𝑬= 𝒐𝒓 𝑬 =𝒌 𝟐 𝟒𝝅𝜺𝟎 𝒓𝟐 𝒓 Example 7 Consider an atom containing a point positive charge Ze at its center and surrounded by a distribution of negative electricity, - Ze uniformly distributed within a sphere of radius R. Prove that the electric field E at a distance r from the center for a point inside the atom is: Solution 𝒁𝒆 𝟏 𝒓 𝑬= 𝟐 − 𝟑 𝟒𝝅𝜺𝟎 𝒓 𝑹 There are two electric fields at this point inside the sphere 1- Eout due to the +ve point charge (Ze) at the center Point charge 𝒒 𝒁𝒆 𝑬𝒐𝒖𝒕 = 𝒌 𝟐 = 𝒓 𝟒𝝅𝜺𝟎 𝒓𝟐 2- Ein due to the –ve distributed charge Ze R Inside a sphere 𝒒𝒓 𝒁𝒆𝒓 r of charges 𝑬𝒊𝒏 =𝒌 𝟑= - Ze 𝑹 𝟒𝝅𝜺𝟎 𝑹𝟑 𝒁𝒆 𝒁𝒆𝒓 𝑬 = 𝑬𝒐𝒖𝒕 − 𝑬𝒊𝒏 ∴𝑬= − 𝒁𝒆 𝟏 𝒓 𝟒𝝅𝜺𝟎 𝒓𝟐 𝟒𝝅𝜺𝟎 𝑹𝟑 ∴𝑬= − 𝟒𝝅𝜺𝟎 𝒓𝟐 𝑹𝟑 Example (2.7) It is found experimentally that the electric field in a certain region of the earth’s atmosphere is directed vertically down. At an altitude of 300 m the field has magnitude 60 N/C, and at an altitude of 200 m, 100 N/C. Find the net amount of charge contained in a cube 100 m on edge, with horizontal faces at altitudes of 200 and 300 m. Neglect the curvature of the earth. Solution Apply the Gauss law on upper and lower surface of cube. ∵ 𝚽𝒖𝒑𝒑𝒆𝒓 = ර𝑬 𝒅𝑨 𝒄𝒐𝒔 𝜽 = 𝑬 ර 𝒅𝑨 𝒄𝒐𝒔𝝅 𝚽𝒖𝒑𝒑𝒆𝒓 = −𝑬𝒖𝒑𝒑𝒆𝒓 𝑨𝒄𝒐𝒔𝟏𝟖𝟎 = −𝟔𝟎 × 𝟏𝟎𝟎𝟐 𝚽𝒍𝒐𝒘𝒆𝒓 = ර𝑬 𝒅𝑨 𝒄𝒐𝒔 𝜽 = 𝑬 ර 𝒅𝑨 𝐜𝐨𝐬 𝟎 𝚽𝒍𝒐𝒘𝒆𝒓 = 𝑬𝑳𝒐𝒘𝒆𝒓 𝑨𝒄𝒐𝒔𝟎 = 𝟏𝟎𝟎 × 𝟏𝟎𝟎𝟐 ∴ 𝑸𝒊𝒏 = 𝜱𝒖𝒑𝒑𝒆𝒓 + 𝜱𝒍𝒐𝒘𝒆𝒓 𝜺𝟎  𝑸𝒊𝒏 ∆𝜱 = 𝜺𝟎 ∴ 𝑸𝒊𝒏 = ∆𝜱𝜺𝟎 ∴ 𝑸𝒊𝒏 = 𝜱𝒖𝒑𝒑𝒆𝒓 + 𝜱𝒍𝒐𝒘𝒆𝒓 𝜺𝟎 ∴ 𝑸𝒊𝒏 = (𝟖. 𝟖𝟓𝟒𝒙𝟏𝟎−𝟏𝟐 )(𝟏𝟎𝟎)𝟐 (𝟏𝟎𝟎 − 𝟔𝟎) ∴ 𝑸𝒊𝒏 = 𝟑. 𝟓𝟒𝝁𝑪  Summary of Laws Spherical Symmetry Electic Flux 𝑸 𝑸𝒓 Inside 𝑬 = 𝒌 Outside 𝑬 = 𝒌 𝑸𝒊𝒏 𝒓𝟐 𝒂𝟑 𝚽= = 𝑬. 𝑨 = 𝑬 𝑨 𝒄𝒐𝒔 𝜽 𝜺𝒐 Cylindrical or axial Symmetry 𝝀 Gauss’ Law 𝑬 = 𝟐𝒌 𝒓 𝑸𝒊𝒏 A planner Symmetry 𝜱= = ර𝑬 ∙ 𝒅𝑨 𝜺𝟎 𝝈 = ‫𝑬 = 𝜽 𝒔𝒐𝒄 𝑨𝒅 𝑬 ׯ‬. 𝑨𝒏𝒐𝒓𝒎𝒂𝒍 Non-conductor sheet ∴ 𝑬 = 𝟐𝜺𝟎 𝝈 Conductor sheet ∴𝑬= 𝜺𝟎  Summary of Laws Co axial Spherers Co axial Cylindrical 𝑸 𝒓