Week 2 Gausss Law general physics 2 pdf d
- 1. GAUSS’S LAW Gauss’s Law an alternative method for calculating electric fields that are produced by highly symmetric charge distributions more conventient and easy Gauss’s law give a relation between the charge and electric flux Electric Flux Let’s consider a uniform electric field lines which are directed along the y axis 𝑬 uniform 𝑨 Then let’s locate a rectangular shaped plate with the surface A which is perpendicular to the electric field lines. Electric field lines penerate the surface area of the plate. The total number of the electric field lines penerating the surface area A is proportional to the product of E and A! 𝜱𝑬 = 𝑬𝑨 known as electric flux that is proportional to the number of the electric field lines penerating perpendicularly to the surface plane 𝑁 𝐶 𝑚2 𝑁𝑚2 𝐶 How electric flux can be a scaler quantity?
- 2. GAUSS’S LAW If the electric field lines do not penerate to the surface plane perpendicularly, how we can write the electric flux? 𝑬 uniform 𝑨 𝜽 • We should find the electric field vector component which is perpendicular to the surface area A! 𝑨 𝜽 𝑬⊥ 𝑬⊥ = 𝑬𝒄𝒐𝒔𝜽 is penerating the surface perpendicularly 𝑨 𝑨 𝜽 𝑨 𝜽 𝑨⊥ 𝑨 is the surface area vector which is perpendicular to surface! 𝑨⊥ = 𝑨𝒄𝒐𝒔𝜽 So 𝑨 is parallel to the 𝑬⊥ or 𝑬 is parallel to the 𝑨⊥! 𝜱𝑬 = 𝑬 ⋅ 𝑨 = 𝑬𝑨 𝒄𝒐𝒔𝜽 Scalar product Scalar quantity Angle between the electric field lines and surface area vector!
- 3. GAUSS’S LAW 𝑨 is always perpendicular to surface area and also directed outward from the closed surface! 𝜱𝑬 = 𝑬 ⋅ 𝑨 = 𝑬𝑨 𝒄𝒐𝒔𝜽 only valid for uniform electric field lines! 𝜱𝑬 = න 𝑬 ⋅ 𝒅𝑨 for non-uniform electric field lines! Differential surface area element, perpendicular to the surface! Exercise: a positive charge of q=3 µC is located at the center of a sphere with radius of 0.2 m. Find the electric flux through the surface of the sphere. +𝑞 𝑅 𝑞 = 3 𝜇𝐶 = 3 10−6𝐶 Let’s define the type of electric field electric field lines on the surface of the sphere is not uniform. WHY? Every point on the sphere surface has the same distance to the point charge q. 𝑬 = 𝒌𝒆𝒒 𝒓𝟐 Magnitude of the E field on the sphere surface 𝑬𝟏 𝒅𝑨𝟏 • q creates an E field directed outward, every point on the surface has same magnitude of electric field but with different direction! Due to this E field is not uniform on the surface! 𝑬𝟐 𝒅𝑨𝟐 𝑬𝟑 𝒅𝑨𝟑
- 4. GAUSS’S LAW +𝑞 𝑅 𝑬𝟏 𝒅𝑨𝟏 𝑬𝟐 𝒅𝑨𝟐 𝑬𝟑 𝒅𝑨𝟑 𝛷𝐸 = ර 𝐸 ⋅ 𝑑𝐴 = ර 𝐸 𝑑𝐴 𝑐𝑜𝑠𝜃 Integral over a closed surface, A! = ර 𝐸 𝑑𝐴 𝑐𝑜𝑠00 There is always 0 degrees between E and dA for this question 𝛷𝐸 = ර 𝐸𝑑𝐴 = 𝐸 ර 𝑑𝐴 Surface area of the sphere Magnitude of the E field is constant! 𝜱𝑬 = 𝑬 𝟒𝝅𝑹𝟐 = 𝑘𝑒𝑞 𝑅2 4𝜋𝑅2 𝒌𝒆 = 𝟏/𝟒𝝅𝜺𝟎 = 𝑞 𝑅2 4𝜋𝑅2 𝟒𝝅𝜺𝟎 𝜱𝑬 = 𝒒 𝜺𝟎 Electric field of the point charge! = 3 10−6𝐶 8.85 10−12𝐶2/𝑁𝑚2 𝑞 = 3 𝜇𝐶 = 3 10−6𝐶 Permittivity of free space 8.8542 10−12 𝐶2 /𝑁𝑚2 = 0.339 106 𝑁𝑚2 /𝐶
- 5. GAUSS’S LAW Let’s consider closed surface in the uniform electric field E as shown! Let’s focus on some small surface area elements! (1) 𝜽 < 𝟗𝟎𝟎 and 𝜱𝑬 > 𝟎 𝜱𝑬 = 𝑬 ⋅ 𝑨 = 𝑬𝑨 𝒄𝒐𝒔𝜽 The electric flux through this A element is positive (2) 𝜽 = 𝟗𝟎𝟎 and 𝑬 ⊥ ∆𝑨𝟐 ; 𝜱𝑬 = 𝟎 (3) 𝜽 > 𝟗𝟎𝟎 and 𝒄𝒐𝒔𝜽 < 𝟎; 𝜱𝑬 < 𝟎 The electric flux through this A element is zero The electric flux through this A element is negative The net flux : number of lines leaving the surface minus the number of lines entering the surface. If more lines are leaving than entering; 𝜱𝑬,𝒏𝒆𝒕 > If more lines are entering than leaving; 𝜱𝑬,𝒏𝒆𝒕 < If leaving and entering lines are same; 𝜱𝑬,𝒏𝒆𝒕 = 𝟎
- 6. 𝒅𝑨𝟔 GAUSS’S LAW Exercise: consider a uniform electric field E oriented in the x direction in empty space. A cube of edge length l is placed in the field as shown. Find the net electric flux through the surface area of the cube. 𝑥 𝑦 𝑧 𝑬 uniform 𝒅𝑨𝟏 𝒍 𝒍 𝒅𝑨𝟑 𝒅𝑨𝟐 𝒅𝑨𝟒 𝒅𝑨𝟓 Cube has six surfaces and all of them same square. Let’s calculate each surface electric flux seperately and then obtain the net electric flux of the cube. 𝛷𝐸,𝑛𝑒𝑡 = 𝑖=1 6 𝛷𝐸,𝑖 𝐸 = 𝐸 Ƹ 𝑖 𝛷𝐸 = න 𝐸 ⋅ 𝑑𝐴 = න 𝐸 𝑑𝐴 𝑐𝑜𝑠𝜃 If 𝜃 = 900, 𝛷𝐸 = 0 For 3,4,5 and 6 surfaces, 𝜱𝑬 = 𝟎 𝛷𝐸,𝑛𝑒𝑡 = 𝛷𝐸,1 + 𝛷𝐸,2 1st surface is placed in the yz plane; 𝑑𝐴 = −𝑑𝐴 Ԧ 𝑖 𝛷𝐸,1 = න 𝐸 Ƹ 𝑖 ⋅ (−𝑑𝐴 Ԧ 𝑖) = −𝐸 න 𝑑𝐴 = −𝐸𝑙2 2nd surface is placed in the yz plane; 𝑑𝐴 = +𝑑𝐴 Ԧ 𝑖 𝛷𝐸,2 = න 𝐸 Ƹ 𝑖 ⋅ (𝑑𝐴 Ԧ 𝑖) = 𝐸 න 𝑑𝐴 = 𝐸𝑙2 𝛷𝐸,𝑛𝑒𝑡 = 0 We obtained a very important result; when any closed surface is located in a uniform electric field the net flux through the closed surface that surrounds no charge is zero! When a charged particle is placed in a closed surface or closed surface is located in a non uniform electric field the net flux can not be zero!
- 7. GAUSS’S LAW Exercise: a triangular prism is located in a uniform E field of 𝐸 = 600 𝑖 𝑁/𝐶. Find the electric flux for each five surfaces of the prism. 𝑬 = 𝟔𝟎𝟎 𝒊 𝑵/𝑪 𝜃 3 𝑚 3 𝑚 5 𝑚 4 𝑚 𝜃 900 − 𝜃 For the front and back triangular surfaces in the xy plane E flux must be zero! 𝑥 𝑦 𝑧 𝒅𝑨𝟐 𝒅𝑨𝟏 𝑑𝐴1 = 𝑑𝐴1 𝑘 𝑑𝐴2 = −𝑑𝐴2 𝑘 𝐸 ⊥ 𝑑𝐴1 𝐸 ⊥ 𝑑𝐴2 𝛷𝐸,1 = 𝛷𝐸,2 = 0 𝜃 𝑥 𝑦 𝑧 𝒅𝑨𝟑 𝑑𝐴3 = −𝑑𝐴3 Ƹ 𝑖 𝛷𝐸,3 = න 𝐸 ⋅ 𝑑𝐴3 = න 600 Ƹ 𝑖 ⋅ −𝑑𝐴3 Ƹ 𝑖 𝛷𝐸,3 = −600 32 = −5400 𝑁𝑚2 /𝐶 𝒅𝑨𝟒 𝑑𝐴4 = −𝑑𝐴4 Ƹ 𝑗 𝐸 ⊥ 𝑑𝐴4 𝛷𝐸,4 = 0 𝒅𝑨𝟓 𝜃 𝛷𝐸,5 = න 𝐸 ⋅ 𝑑𝐴5 = න 𝐸𝑑𝐴5𝑐𝑜𝑠𝜃 𝛷𝐸,5 = 600 3 5 3 5 = 5400 𝑁𝑚2/𝐶 𝛷𝐸,𝑛𝑒𝑡 = 0
- 8. GAUSS’S LAW Exercise: a closed surface with dimensions of a=0.2 m, b=0.3 m and c=0.3 m is located in the electric field of 𝐸 = 3 + 2𝑥2 Ԧ 𝑖 𝑁/𝐶 where x is in meters. Calculate the net electric flux leaving the closed surface. 𝑥 𝑦 𝑧 𝑥 = 𝑎 𝑐 𝑎 𝑏 The electric field is not uniform it varies with x coordinate. The electric flux must be calculated by using integral! 𝒅𝑨𝟏 𝒅𝑨𝟐 Non zero electric flux is observed for the 1st and 2nd surfaces. 𝑑𝐴1 = −𝑑𝐴1 Ƹ 𝑖 𝛷𝐸,1 = න 𝐸1 ⋅ 𝑑𝐴1 = න(3 + 2𝑎2) Ƹ 𝑖 ⋅ −(𝑑𝐴1) Ƹ 𝑖 = (3 + 2𝑎2) Ƹ 𝑖 ⋅ −(𝑎𝑏) Ƹ 𝑖 = −(3 + 2𝑎2)𝑎𝑏 𝑑𝐴2 = 𝑑𝐴2 Ƹ 𝑖 𝛷𝐸,2 = න 𝐸2 ⋅ 𝑑𝐴2 = න(3 + 2(𝑎 + 𝑐)2) Ƹ 𝑖 ⋅ (𝑑𝐴2) Ƹ 𝑖 = ((3 + 2(𝑎 + 𝑐)2 )) Ƹ 𝑖 ⋅ (𝑎𝑏) Ƹ 𝑖 = (3 + 2𝑎2 + 4𝑎𝑐 + 2𝑐2 )𝑎𝑏 𝛷𝐸,𝑛𝑒𝑡 = − 3 + 2𝑎2 𝑎𝑏 + (3 + 2𝑎2 + 4𝑎𝑐 + 2𝑐2)𝑎𝑏 𝛷𝐸,𝑛𝑒𝑡 = − 3𝑎𝑏 + 2𝑎3 𝑏 + (3𝑎𝑏 + 2𝑎3 𝑏 + 4𝑎2 𝑏𝑐 + 2𝑎𝑏𝑐2 ) 𝛷𝐸,𝑛𝑒𝑡 = 4𝑎2𝑏𝑐 + 2𝑎𝑏𝑐2 = 4 0.22 0.3 0.3 + 2 0.2 0.3 0.32 = 0.2688 𝑁𝑚2/𝐶
- 9. GAUSS’S LAW Gauss’s Law Gauss’s Law relates electric charge and electric field a more general and elegant form of Coulumb’s law. The net electric flux through any closed surface is given by 𝒒𝒊𝒏/𝜺𝟎 where; «𝒒𝒊𝒏» equals the net charge inside the closed surface known as Gaussian surface. 𝜱𝑬 = ර 𝑬 ⋅ 𝒅𝑨 = 𝒒𝒊𝒏 𝜺𝟎 Gauss’s Law The electric field at any point on the Gaussian surface Closed surface The electric field created by highly symmetric charge distributions can be calculated easily by using Gauss’s law. The way to follow when Gauss’s law is applied: 1) A closed Gaussian surface is drawn in accordance with symmetry of the charge distribution 2) The point where electric field will be calculated must coincide with choosen Gaussian surface 3) The electric field vector 𝑬 and the surface area vector 𝒅𝑨 are drawn and the electric flux is written equal to 𝒒𝒊𝒏/𝜺𝟎 . Hence the electric field is calculated. ❖ before we have calculated the electric flux of a +q point charge through a spherical surface. 𝜱𝑬 = 𝒒 𝜺𝟎 for spherical Gaussian surface
- 10. GAUSS’S LAW The net electric flux is the same through all surfaces. For 𝑺𝟐 and 𝑺𝟑 non spherical surfaces; the net electric flux does not change. Because the net electric flux is proportional to the number of electric field lines passing through a surface. As shown, number of lines passing through all surfaces are equal. The net flux through any closed surface surrounding a point charge q is given by 𝒒/𝜺𝟎 and it is independent of shape of that surface. In this condition, the point charge is outside the closed surface. So the number of electric field lines entering and leaving the closed surface are the same. 𝜱𝑬 = ර 𝑬 ⋅ 𝒅𝑨 = 𝒒𝒊𝒏 𝜺𝟎 = 𝟎 𝜱𝑬,𝑺 = 𝒒𝟏 𝜺𝟎 𝜱𝑬,𝑺′ = 𝒒𝟐+ 𝒒𝟑 𝜺𝟎 𝜱𝑬,𝑺′′ = 𝟎
- 11. GAUSS’S LAW Exercise: Find the electric field created by an isolated point charge of q at r distance far from the charge by using Gauss’s law. Spherical Gaussian surface is suitable closed surface that surrounds the point charge of q. You should choose a Gaussian surface such that the magnitude of the electric field at each point on the surface must be the same. Gaussian surface is an imaginary surface. When the charge is at the center of the sphere the electric field is everywhere normal to the surface and constant in magnitude. The electric field’s magnitude is constant at every point on the Gaussian surface, because there is a constant r distance between the charge and the point on the surface. For every point on the sphere 𝑬 ∕∕ 𝒅𝑨, 𝜃 = 00 𝛷𝐸 = ර 𝐸 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 ර 𝐸𝑑𝐴𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 𝐸 ර 𝑑𝐴 = 𝐸4𝜋𝑟2 = 𝒒 𝜀0 𝐸 = 𝑞 𝟒𝝅𝑟2𝜺𝟎 𝑬 = 𝒌𝒆 𝒒 𝒓𝟐 ො 𝒓 The unit vector which is radially directed outward! ො 𝒓
- 12. GAUSS’S LAW Exercise: a spherically symmetric charge distribution. An insulating solid sphere of radius a has a uniform volume charge density, 𝜌 and carries a total positive charge Q. a) Calculate the magnitude of the electric field at a point outside the sphere b) Find the magnitude of the electric field at a point inside the sphere. 𝒂 𝝆 𝑸 For points outside the sphere a large spherical gaussian surface is drawn concentric with the sphere. 𝑟 > 𝑎 ර 𝐸1 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 𝑑𝐴 𝐸 𝑑𝐴 𝐸 𝑑𝐴 𝐸 For each point on the Gaussian surface 𝐸 ∕∕ 𝑑𝐴, 𝜃 = 00 ර 𝐸1𝑑𝐴 𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 = 𝑄 𝜀0 How much charge is trapped in the Gaussian surface ? = 𝐸1 4𝜋𝑟2 𝐸1 = 𝑘𝑒𝑄/𝑟2 For points inside the sphere, a spherical Gaussian surface smaller than the sphere is drawn. 𝑑𝐴 𝐸 ර 𝐸2 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 ර 𝐸2𝑑𝐴 𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 𝐸24𝜋𝑟2 = 𝑞𝑖𝑛 𝜀0 This time net charge inside the Gaussian surface is <Q!
- 13. GAUSS’S LAW Since the charge distribution is uniform q inside can be determined by constructing a direct proportion. The total charge 𝑄 is distributed over volume 4 3 𝜋𝑎3 Charge of 𝑞𝑖𝑛 distributed over volume 4 3 𝜋𝑟3 Q 4 3 𝜋𝑟3 = 𝑞𝑖𝑛 4 3 𝜋𝑎3 Q 𝑟3 /𝑎3 = 𝑞𝑖𝑛 𝐸24𝜋𝑟2 = 𝑞𝑖𝑛 𝜀0 = Q 𝑟3 𝜀0𝑎3 𝐸2 = 𝑘𝑒𝑄𝑟/𝑎3 𝑟 = 𝑎, is critical distance 𝐸1 = 𝐸2 𝐸1(𝑎) = 𝑘𝑒𝑄/𝑎2 𝐸2(𝑎) = 𝑘𝑒𝑄𝑎/𝑎3 𝒂 What if the sphere was a conductor?
- 14. GAUSS’S LAW Exercise: a solid insulating sphere of radius a has a non-uniform charge density of 𝜌 = 𝛼𝑟2 where 𝛼 is a positive constant and r is the radial distance from the center of the sphere. a) Find the 𝛼 constant in terms of the total charge of the sphere, Q b) Find the electric field outside the sphere c) Find the electric field inside the sphere 𝒂 𝝆 = 𝜶𝒓𝟐 The charge distribution is non uniform, hence the charge will be calculated by using integral not a direct proportion! 𝑄 = න 𝑑𝑞 = න 𝝆𝑑𝑉 = න 𝜶𝒓𝟐 𝟒𝝅𝒓𝟐 𝒅𝒓 = 𝑉𝑠𝑝ℎ𝑒𝑟𝑒 = 4 3 𝜋𝑟3 𝑑𝑉𝑠𝑝ℎ𝑒𝑟𝑒 = 4 3 𝜋3𝑟2𝑑𝑟 4𝜋𝛼 න 𝑟=0 𝑟=𝑎 𝑟4 𝑑𝑟 𝑄 = 4𝜋𝛼 𝑟5 5 = 4 5 𝜋𝛼𝑎5 𝑑𝑉𝑠𝑝ℎ𝑒𝑟𝑒 = 𝟒𝝅𝒓𝟐𝒅𝒓 Surface area x thickness 𝛼 = 5𝑄 4𝜋𝑎5 𝒂 𝑟 > 𝑎 ර 𝐸1 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 𝑑𝐴 ර 𝐸1𝑑𝐴 𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 = 𝑄 𝜀0 = 𝐸1 4𝜋𝑟2 𝐸1 = 𝑘𝑒𝑄/𝑟2
- 15. GAUSS’S LAW 𝒂 𝑑𝐴 𝐸 𝑟 < 𝑎, ර 𝐸2 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 ර 𝐸2𝑑𝐴 𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 𝑞𝑖𝑛= 𝑑𝑞 = 𝝆𝑑𝑉 = 4𝜋𝛼 𝑟=0 𝑟=𝑟 𝑟4 𝑑𝑟 = 4𝜋𝜶𝑟5 5 = 4𝜋𝑟5 5 5𝑄 4𝜋𝑎5 = 𝑄𝑟5 𝑎5 𝐸24𝜋𝑟2 = 𝑄𝑟5 𝑎5 𝜀0 𝐸2 = 𝑄𝑟3 𝑎5 4𝜋𝜀0 = 𝑘𝑒𝑄𝑟3 𝑎5 𝑟
- 16. GAUSS’S LAW Exercise: Find the electric field at a distance r from a wire with positive charge, infinite length and constant charge per unit length, 𝜆. +∞ −∞ 𝒓 + + + + The must suitable Gaussian Surface (GS) that surrounds the linear charge density is cylindrical GS with radius r and length l. 𝑷 𝒓 𝑷 + + + + 𝒍 𝒅𝑨𝟏 𝒅𝑨𝟐 𝑬𝟐 𝒅𝑨𝟑 𝑬𝟏 𝑬𝟑 Top view + 𝑬 ර 𝐸 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 We have two flat ends of cylinder and the curved lateral surface! = න 𝐸1 ⋅ 𝑑𝐴1 + න 𝐸2 ⋅ 𝑑𝐴2 + න 𝐸3 ⋅ 𝑑𝐴3 𝐸1 ⊥ 𝑑𝐴1 𝐸3 ⊥ 𝑑𝐴3 The field magnitude is same for all points on lateral surface and E is always parallel to dA. 𝑞𝑖𝑛 𝜀0 = න 𝐸2 ⋅ 𝑑𝐴2 = න 𝐸2𝑑𝐴2𝑐𝑜𝑠00 𝐸22𝜋𝑟𝒍 = 𝑞𝑖𝑛 𝜀0 𝑞𝑖𝑛: the charge surrounded by the cylinder with radius r and length l 𝑞𝑖𝑛= 𝜆𝑙 𝐸22𝜋𝑟𝑙 = 𝜆𝑙 𝜀0 𝑬𝟐 = 𝟐𝒌𝒆𝝀 𝒓 Compare with previous week’s results!
- 17. GAUSS’S LAW Exercise: Find the electric field due to an infinite plane (insulator*) of positive charge with uniform surface charge density, 𝜎. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Similar to our previous question we can select a cylindrical GS that includes small portions of the charge distribution. ර 𝐸 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 = න 𝐸1 ⋅ 𝑑𝐴1 + න 𝐸2 ⋅ 𝑑𝐴2 + න 𝐸3 ⋅ 𝑑𝐴3 𝒅𝑨𝟏 𝑬𝟏 𝒅𝑨𝟐 𝑬𝟐 𝒅𝑨𝟑 𝑬𝟑 𝐸2 ⊥ 𝑑𝐴2 𝑞𝑖𝑛 𝜀0 = න 𝐸1𝑑𝐴1𝑐𝑜𝑠00 + න 𝐸3𝑑𝐴3𝑐𝑜𝑠00 The electric field’s magnitude will be same for the right and left ends. Because the distances between the charge distribution and the ends are the same! 𝑞𝑖𝑛 𝜀0 = 2𝐸𝐴 𝒒𝒊𝒏= 𝝈𝑨 𝜎𝑑𝐴 𝜀0 = 2𝐸𝑑𝐴 𝑬 = 𝝈 𝟐𝜺𝟎 Compare with previous results!
- 18. GAUSS’S LAW Conductors in electrostatic equilibrium we have focused on the calculation of electric fields due to insulating materials. The charge carriers can not move in the insulators but they can freely move within the conducting materials. When there is no net motion of charge within a conductor the conductor said to be in electrostatic equilibrium. In gauss problems we will often face with the conductors in electrostatic equilibrium. A conductor in electrostatic equilibrium must satisfy the following properties; 1) The electric field is zero everywhere inside the conductor whether conductor is solid or hollow • Let’s locate a conducting slab in an external electric field E The free electrons accelerates to the left and the motion of the electrons causes a positively charged right surface. So an opposite electric field to the external electric field E is established inside the conductor. The motion of the electrons continues until the electric field inside conductor balances an external electric field vector E.
- 19. GAUSS’S LAW 1) The electric field is zero everywhere inside the conductor whether conductor is solid or hollow The net electric field inside the conductor becomes zero in ~ 10-16 s for good conductors. Additionally if electric field were not zero inside the conductor the free electrons would experience an eE electric forces on themselves and they would accelerate. 2) If an isolated conductor carries a charge the charge resides on its surface. GS very close to the conductor surface! + + + + + + + + + + + + + E=0 The net charge inside GS must be zero because the E field is zero inside the conductor that is at electrostatic equilibrium. ර 𝐸 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 if conductor is charged these charges must reside on its surface otherwise net zero electric field condition are not satisfied.
- 20. GAUSS’S LAW 3) The electric field just outside a charged conductor is perpendicular to the surface and it has a magnitude of 𝜎/𝜀0 where sigma «𝜎» is the surface charge density at that point. 𝐸𝐴 small cylindircal GS is choosen Some part of the cylinder is outside and remaining part is inside the condutor Let’s consider + conducting material. The non zero electric flux comes from only the flat surface which is outside conductor. The non zero electric flux comes from only the flat surface which is outside conductor. Because the electric field inside the conductor is zero, so that surface inside the conductor does not experience any electric field lines and also electric flux. For the lateral surface E is parallel to surface and there is always 90° between E and dA. 𝛷𝐸 = ර 𝐸 ⋅ 𝑑𝐴 = න 𝒐𝒖𝒕𝒔𝒊𝒅𝒆 𝒇𝒍𝒂𝒕 𝒔𝒖𝒓𝒇𝒂𝒄𝒆 . 𝑬 ⋅ 𝒅𝑨 + න 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸 ⋅ 𝑑𝐴 + න𝑖𝑛𝑠𝑖𝑑𝑒 𝑓𝑙𝑎𝑡 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 . 𝐸 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 න 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑓.𝑠. . 𝐸𝑑𝐴 𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 𝐸𝐴 = 𝑞𝑖𝑛 𝜀0 = 𝜎𝐴 𝜀0 𝑬 = 𝝈 𝜺𝟎 The magnitude of the electric field immediately outside a charged conductor.
- 21. GAUSS’S LAW 4) On a irregularly shaped conductor the surface charge density is greatest at locations where the radius of the curvature of the surface is smallest. + increasing curvarture decreasing curvarture radius flat + + + + + + + + Exercise: a solid insulating sphere of radius R has a non uniform charge density, 𝜌 = 𝛼𝑟. And has a total charge of +2Q (where alpha is a positive constant and r is the radial distance from the origin) concentric with the sphere, a conducting spherical shell with the inner and outer radii of 2R and 3R carries +4Q. a) Find the alpha constant in terms of Q and R b) Find the magnitude of the electric field in the regions 𝑹 𝟐𝑹 +𝟐𝑸 +𝟒𝑸 𝟑𝑹 To determine 𝛼 we can write the total charge of the insulating sphere 2𝑄 = න 𝑑𝑞 = න 𝜌𝑑𝑉 𝑉 = 4 3 𝜋𝑟3 𝑑𝑉 = 4𝜋𝑟2𝑑𝑟 2𝑄 = න 𝑟=0 𝑟=𝑅 𝛼𝑟4𝜋𝑟2𝑑𝑟
- 22. GAUSS’S LAW 2𝑄 = 4𝜋𝛼 න 𝑟=0 𝑟=𝑅 𝑟3𝑑𝑟 = 4𝜋𝛼 𝑟4 4 = 𝜋𝛼𝑅4 𝛼 = 2𝑄/𝜋𝑅4 𝑅 2𝑅 +𝟐𝑸 +4𝑄 3𝑅 𝒓 𝑑𝐴 𝐸1 𝒓 𝐸1 ∕∕ 𝑑𝐴 ර 𝐸1 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 𝑞𝑖𝑛= 𝜌𝑑𝑉 = 𝑟=0 𝑟=𝑟 𝛼𝑟4𝜋𝑟2 𝑑𝑟 = 4𝜋𝛼 න 𝑟=0 𝑟=𝑟 𝑟3𝑑𝑟 = 𝜋𝛼𝑟4 𝑞𝑖𝑛= 𝜋 2𝑄 𝜋𝑅4 𝑟4 = 2𝑄 𝑟 𝑅 4 𝐸1 ර 𝑑𝐴 = 𝐸14𝜋𝑟2 = 2𝑄 𝑟 𝑅 4 𝜀0 𝐸1 = 2𝑄 𝑟 𝑅 4 4𝜋𝑟2𝜀0 = 𝑘𝑒 2𝑄𝑟2 𝑅4 For r<R +𝟐𝑸 For R<r<2R 𝟐𝑹 ර 𝐸2 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 𝐸2 = 2𝑄 4𝜋𝑟2𝜀0 = 𝑘𝑒 2𝑄 𝑟2 𝐸2 ∕∕ 𝑑𝐴 𝑑𝐴 𝐸2 𝒓 𝐸3 For 2R<r<3R region corresponds to the inside of the conducting spherical shell! According to electrostatic equilibrium condition; 𝐸3 = 0
- 23. GAUSS’S LAW 2𝑅 +𝟐𝑸 +4𝑄 3𝑅 𝒓 𝑑𝐴 𝐸4 For r>3R ර 𝐸4 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 𝐸4 = 6𝑄 4𝜋𝑟2𝜀0 = 𝑘𝑒 6𝑄 𝑟2 𝐸4 ∕∕ 𝑑𝐴 +𝟐𝑸 −𝟐𝑸 ∶ inner surface +𝟔𝑸 ∶ outer surface
- 24. GAUSS’S LAW Exercise: a solid insulating sphere of radius a has a non uniform charge density 𝜌 = −𝐴𝑟 where A is a positive constant and r is the radial distance from the origin concentric with the sphere, a neutral conducting spherical shell with the inner and outer radii are of b and c is located a) Find the magnitude of the electric field in the regions b) Determine the induced charge densities on the inner and outer surfaces of the conductive shell. 𝑎 𝑏 𝑐 𝑑𝐴 𝐸1 𝒓 𝑟 < 𝑎; 𝐸1 ∕∕ 𝑑𝐴 ර 𝐸1𝑑𝐴𝑐𝑜𝑠1800 = 𝑞𝑖𝑛 𝜀0 𝑞𝑖𝑛= 𝜌𝑑𝑉 = 𝑟=0 𝑟=𝑟 −𝐴𝑟4𝜋𝑟2 𝑑𝑟 = −4𝜋𝐴 න 𝑟=0 𝑟=𝑟 𝑟3 𝑑𝑟 = −𝜋𝐴𝑟4 𝐸1 ර 𝑑𝐴 = 𝐸14𝜋𝑟2 = −𝜋𝐴𝑟4 𝜀0 𝐸1 = − 𝐴𝑟2 4𝜀0 𝐸1 = − 𝐴𝑟2 4𝜀0 Ƹ 𝑟
- 25. GAUSS’S LAW 𝑎 𝑏 𝑐 𝑑𝐴 𝐸2 𝒓 𝑎 < 𝑟 < 𝑏; 𝐸2 ∕∕ 𝑑𝐴 ර 𝐸2𝑑𝐴𝑐𝑜𝑠1800 = 𝑞𝑖𝑛 𝜀0 𝑞𝑖𝑛= 𝜌𝑑𝑉 = 𝑟=0 𝑟=𝑎 −𝐴𝑟4𝜋𝑟2 𝑑𝑟= −4𝜋𝐴 න 𝑟=0 𝑟=𝑟 𝑟3 𝑑𝑟 = −𝜋𝐴𝑎4 𝐸2 ර 𝑑𝐴 = 𝐸24𝜋𝑟2 = −𝜋𝐴𝑎4 𝜀0 𝐸2 = − 𝐴𝑎4 4𝜀0𝑟2 𝐸2 = − 𝐴𝑎4 4𝜀0𝑟2 Ƹ 𝑟 𝐸3 = 0 𝑏 < 𝑟 < 𝑐; 𝐸3 = 0 The electric field should be zero inside the conducting sheel! 𝑟 > 𝑐; 𝐸4 ∕∕ 𝑑𝐴 𝑎 𝑏 𝑐 𝑑𝐴 𝐸4 𝒓 ර 𝐸4𝑑𝐴𝑐𝑜𝑠1800 = 𝑞𝑖𝑛 𝜀0 𝑞𝑖𝑛= −𝜋𝐴𝑎4
- 26. GAUSS’S LAW 𝐸4 ර 𝑑𝐴 = 𝐸44𝜋𝑟2 = −𝜋𝐴𝑎4 𝜀0 𝐸4 = − 𝐴𝑎4 4𝜀0𝑟2 𝐸4 = − 𝐴𝑎4 4𝜀0𝑟2 Ƹ 𝑟 𝑎 𝑏 𝑐 conducting sheel is in electrostatic equilibrium! 𝑟 = 𝑏 inner surface 𝑟 = 𝑐 outer surface Let’s write the gauss’s law for inner surface; as known 𝑟 = 𝑏, 𝐸 = 0. 𝛷𝐸 = ර 𝐸 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 ර 0 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 0 = 𝑞𝑖𝑛= 𝑞𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑛𝑔 𝑠𝑝ℎ𝑒𝑟𝑒 + 𝑞 𝑖𝑛𝑛𝑒𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑠ℎ𝑒𝑙𝑙 −𝜋𝐴𝑎4 0 = −𝜋𝐴𝑎4 + 𝑞𝑖𝑠𝑐𝑠 𝒒𝒊𝒔𝒄𝒔= 𝝅𝑨𝒂𝟒 𝜋𝐴𝑎4 −𝜋𝐴𝑎4 Since the conductor sheel is neutral; 𝑞 𝑜𝑢𝑡𝑒𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑠ℎ𝑒𝑙𝑙 + 𝑞 𝑖𝑛𝑛𝑒𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑠ℎ𝑒𝑙𝑙 = 0 𝒒𝒐𝒔𝒄𝒔= −𝝅𝑨𝒂𝟒 −𝜋𝐴𝑎4 𝑞𝑖𝑠𝑐𝑠= 𝜎𝑖𝑠𝑐𝑠𝑑𝐴 = 𝜎𝑖𝑠𝑐𝑠 𝑑𝐴 = 𝜎𝑖𝑠𝑐𝑠4𝜋𝑏2 = 𝝅𝑨𝒂𝟒 𝜎𝑖𝑠𝑐𝑠 = 𝐴𝑎4 /4𝑏2 𝑞𝑜𝑠𝑐𝑠= 𝜎𝑜𝑠𝑐𝑠𝑑𝐴 = 𝜎𝑜𝑠𝑐𝑠 𝑑𝐴 = 𝜎𝑜𝑠𝑐𝑠4𝜋𝑐2 = 𝝅𝑨𝒂𝟒 𝜎𝑜𝑠𝑐𝑠 = 𝐴𝑎4 /4𝑐2
- 27. GAUSS’S LAW Exercise: a right circular cone of radius R and height h is placed with its axis parallel to a uniform electric field E. The direction of the electric field lines is in the –y direction. Determine the electric flux entering the curved surface of the cone. 𝑅 𝑬 Since there is no charge inside closed surface of the cone electric flux must be zero! 𝛷𝐸 = 𝛷𝐸,𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 + 𝛷𝐸,𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = 0 𝛷𝐸,𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = −𝛷𝐸,𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝛷𝐸,𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = න 𝐸 ⋅ 𝑑𝐴 = 𝐸𝑑𝐴 𝑐𝑜𝑠00 = 𝐸 𝑑𝐴 = 𝐸𝜋𝑟2 𝑑𝐴 𝐸 𝛷𝐸,𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = 𝐸𝜋𝑟2 𝛷𝐸,𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = −𝐸𝜋𝑟2
- 28. GAUSS’S LAW Exercise: a point charge of +2Q is located at the center of a spherical insulating shell whose the inner and outer radii are R and 2R. The shell carries a +Q charge and has a uniform charge density 𝜌. Find the electric field in all regions. What if the shell is conducting? +2𝑄 +𝑄 𝑅 2𝑅 𝑅 < 𝑟 < 2𝑅; 𝐸1 ∕∕ 𝑑𝐴 𝑑𝐴 𝐸1 𝒓 ර 𝐸1𝑑𝐴𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 𝑞𝑖𝑛= 𝑞𝑝𝑜𝑖𝑛𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 + 𝑞𝑖𝑛𝑠 𝑠𝑝ℎ𝑒𝑟𝑒 +2𝑄 Homogeneous charge distribution, between r and R 𝑟 < 𝑅; 𝑬𝟎 = 𝟐𝒌𝒆𝑸/𝒓𝟐 𝑉 = 4 3 𝜋𝑟3 = 4 3 𝜋 2𝑅 3 − 𝑅 3 = 4𝜋 3 7𝑅3 Total Q is distributed over the V For 𝑅 < 𝑟 < 2𝑅; 𝑉 = 4 3 𝜋 𝑟 3 − 𝑅 3 = 4𝜋 3 (𝑟3 − 𝑅3) 𝑞𝑖𝑛𝑠 𝑠𝑝ℎ𝑒𝑟𝑒= 𝑄 4𝜋 3 (𝑟3 − 𝑅3) 4𝜋 3 (7𝑅3) = 𝑄 (𝑟3 − 𝑅3 ) (7𝑅3) ර 𝐸1𝑑𝐴𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 = 1 𝜀0 2𝑄 + 𝑄 𝑟3 − 𝑅3 7𝑅3 𝐸14𝜋𝜀0𝑟2 = 2𝑄 + 𝑄 (𝑟3 − 𝑅3) (7𝑅3) 𝐸1 = 14𝑄𝑅3 + 𝑄(𝑟3 − 𝑅3) (7𝑅3)4𝜋𝜀0𝑟2 = 𝑘𝑒 13𝑄𝑅3 + 𝑄𝑟3 7𝑅3𝑟2
- 29. GAUSS’S LAW 𝑬𝟏 = 𝒌𝒆𝑸 𝟕 𝟏𝟑 𝒓𝟐 + 𝒓 𝑹𝟑 +2𝑄 +𝑄 𝑅 2𝑅 For 𝑟 > 2𝑅; 𝑞𝑖𝑛= 2𝑄 + 𝑄 𝒓 𝑑𝐴 𝐸2 ර 𝐸2𝑑𝐴𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 = 3𝑄 𝜀0 = 4𝜋𝑟2 𝑬𝟐 = 𝟑𝒌𝒆𝑸 𝒓𝟐 If the shell is conducting 𝑬𝟏 = 𝟎 because q inside the conductor is zero. 𝑬𝟐 remains same because total charge is constant! +2𝑄 −𝟐𝑸 +𝟑𝑸
- 30. GAUSS’S LAW Exercise: a long straight wire with linear charge density of 𝜆 is located in the xy plane that makes a 450 with the x axis as shown. The charge is distributed along the wire uniformly. a) Find the electric field vector at the point P(0,2d) by using Gauss’s law b) Find the electric force vector acting on a charge of –Q which is located at point P. 0 𝑥 𝑦 𝑷 𝟎, 𝟐𝒅 ∗ +𝝀 450 We should find a GS which P point must coincide on the GS. Cylindrical Gaussian surface fits well 𝒉 𝒓 Every point on the lateral surface of the Gaussian surface including P point must have the same electric field magnitude. 𝛷𝐸 = ර 𝐸 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 𝛷𝐸 = ර 𝐸 ⋅ 𝑑𝐴 = 𝑞𝑖𝑛 𝜀0 = න 𝑢𝑝𝑝𝑒𝑟 𝑓𝑙𝑎𝑡 . … + න 𝑙𝑜𝑤𝑒𝑟 𝑓𝑙𝑎𝑡 . … + න 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 . … 𝑑𝐴1 𝐸1 𝑑𝐴2 𝐸2 𝑑𝐴3 𝐸3 න 𝐸3 ⋅ 𝑑𝐴3 = 𝑞𝑖𝑛 𝜀0 න 𝐸𝑃𝑑𝐴𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 = 𝜆ℎ 𝜀0 𝐸𝑃 න 𝑑𝐴 = 𝜆ℎ 𝜀0 𝑷 𝟎, 𝟐𝒅 ∗ 𝑬𝑷 2𝑑 𝑟 450 450 𝐸𝑃2𝜋𝑟ℎ = 𝜆ℎ 𝜀0
- 31. GAUSS’S LAW 𝐸𝑃 = 𝜆ℎ 2𝜋𝑟ℎ𝜀0 = 𝜆 2𝜋𝑟𝜀0 𝑠𝑖𝑛450 = 𝑟 2𝑑 = 2 2 𝑟 = 𝑑 2 𝑬𝑷 = 𝝀 𝟐𝝅𝜺𝟎 𝟏 𝒅 𝟐 𝑬𝑷 = 𝑬𝑷(−𝒄𝒐𝒔𝟒𝟓𝟎 Ƹ 𝒊 + 𝒔𝒊𝒏𝟒𝟓𝟎 Ƹ 𝒋) 0 𝑥 𝑦 𝑷 +𝝀 450 𝒉 𝒓 −𝑄 𝑭𝒆 𝑬𝑷 𝐸𝑃 = 𝜆 2𝜋𝜀0 1 𝑑 2 − 2 2 Ƹ 𝑖 + 2 2 Ƹ 𝑗 𝐸𝑃 = 𝜆 4𝜋𝜀0 1 𝑑 − Ƹ 𝑖 + Ƹ 𝑗 𝐹𝑒 = 𝑞𝐸 =−𝑄𝐸𝑃 = 𝜆 4𝜋𝜀0 𝑄 𝑑 Ƹ 𝑖 − Ƹ 𝑗
- 32. 𝑹 𝒓 𝒉 GAUSS’S LAW Exercise: an infinitely long insulating cylinder of radius R has a volume charge density that varies with the radius 𝜌 = 𝜌0 𝑎 − 𝑟 𝑏 where all are positive constants and r is the distance from the axis of the cylinder. Use Gauss’s law to determine the magnitude of the electric field at radial distances a) r<R b) r>R 𝑑𝐴1 𝐸1 𝑑𝐴2 𝐸2 𝐸3 𝑑𝐴3 𝑉 = 𝜋𝑟2 ℎ 𝒅𝑽 = 𝟐𝝅𝒓𝒅𝒓𝒉 = 𝒅𝑨𝒅𝒓 𝛷𝐸 = ර 𝐸3 ⋅ 𝑑𝐴3 = 𝑞𝑖𝑛 𝜀0 න 𝐸3 ⋅ 𝑑𝐴3 = න 𝐸𝑑𝐴𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 𝐸 න 𝑑𝐴 = 𝜌𝑑𝑉 𝜀0 𝐸2𝜋𝑟ℎ = 𝜌0 𝑎 − 𝑟 𝑏 (2𝜋𝑟ℎ 𝑑𝑟) 𝜀0 = 2𝜋ℎ𝜌0 0 𝑟 𝑎 − 𝑟 𝑏 𝑟𝑑𝑟 𝜀0 𝐸 = 2𝜋ℎ𝜌0 0 𝑟 𝑎 − 𝑟 𝑏 𝑟𝑑𝑟 2𝜋𝑟ℎ𝜀0 = 𝜌0 0 𝑟 𝑎 − 𝑟 𝑏 𝑟𝑑𝑟 𝑟𝜀0 = 𝜌0 𝑎𝑟2 2 − 𝑟3 3𝑏 𝑟𝜀0 = 𝜌0𝑟 𝜀0 𝑎 2 − 𝑟 3𝑏 For 𝑟 < 𝑅;
- 33. 𝑹 𝒓 𝒉 GAUSS’S LAW 𝑑𝐴1 𝐸1 𝑑𝐴2 𝐸2 𝐸3 𝑑𝐴3 For 𝑟 > 𝑅; 𝛷𝐸 = ර 𝐸3 ⋅ 𝑑𝐴3 = 𝑞𝑖𝑛 𝜀0 න 𝐸3 ⋅ 𝑑𝐴3 = න 𝐸𝑑𝐴𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 𝐸 න 𝑑𝐴 = 𝜌𝑑𝑉 𝜀0 𝐸2𝜋𝑟ℎ = 𝜌0 𝑎 − 𝑟 𝑏 (2𝜋𝑟ℎ 𝑑𝑟) 𝜀0 = 2𝜋ℎ𝜌0 0 𝑅 𝑎 − 𝑟 𝑏 𝑟𝑑𝑟 𝜀0 𝐸 = 2𝜋ℎ𝜌0 0 𝑅 𝑎 − 𝑟 𝑏 𝑟𝑑𝑟 2𝜋𝑟ℎ𝜀0 = 𝜌0 0 𝑅 𝑎 − 𝑟 𝑏 𝑟𝑑𝑟 𝑟𝜀0 = 𝜌0 𝑎𝑅2 2 − 𝑅3 3𝑏 𝑟𝜀0 𝐸 = 𝜌0𝑅2 𝜀0𝑟 𝑎 2 − 𝑅 3𝑏
- 34. GAUSS’S LAW Exercise: an infinitely long straight wire with the uniform linear charge density 𝜆 is lying at the center of the cylindrical long insulating shell with radius R. The cylindrical shell carries uniform surface charge density +𝜎. a) Find the electric field in the region r>R where r is the radial distance from the linear charge distribution b) A point charge of q and mass of m is released at r=R/2. find the acceleration of the point charge when it is released. +𝜎 +𝜆 𝑹 For 𝑟 > 𝑅; 𝛷𝐸 = ׯ 𝐸3 ⋅ 𝑑𝐴3 = 𝑞𝑖𝑛/𝜀0 𝒓 𝒍 𝑑𝐴1 𝐸1 𝑑𝐴2 𝐸2 𝐸3 𝑑𝐴3 න 𝐸3 ⋅ 𝑑𝐴3 = න 𝐸𝑑𝐴𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 𝐸2𝜋𝑟𝑙 = 𝜆𝑑𝑙 + 𝜎𝑑𝐴 𝜀0 𝐴 = 2𝜋𝑅𝑙 𝐸 = 𝜆𝑙 + 𝜎2𝜋𝑅𝑙 2𝜋𝑟𝑙𝜀0 = 𝜆 + 𝜎2𝜋𝑅 2𝜋𝑟𝜀0 𝑞𝑖𝑛= 𝑞𝑙𝑖𝑛𝑒𝑎𝑟 𝑐ℎ𝑎𝑟𝑔𝑒 + 𝑞𝑐𝑦𝑙𝑖𝑛.𝑠ℎ𝑒𝑙𝑙 𝑞𝑖𝑛= 𝜆𝑙 + 𝜎2𝜋𝑅𝑙 𝐸 = 1 𝑟 𝜆 2𝜋𝜀0 + 𝜎𝑅 𝜀0 Since point charge is located at r=R/2 we should find the E field at r=R/2
- 35. GAUSS’S LAW +𝜎 +𝜆 𝑹 𝒓 𝒍 𝑑𝐴1 𝐸1 𝑑𝐴2 𝐸2 𝐸3 For 𝑟 < 𝑅; 𝛷𝐸 = ׯ 𝐸3 ⋅ 𝑑𝐴3 = 𝑞𝑖𝑛/𝜀0 න 𝐸3 ⋅ 𝑑𝐴3 = න 𝐸𝑑𝐴𝑐𝑜𝑠00 = 𝑞𝑖𝑛 𝜀0 𝐸2𝜋𝑟𝑙 = 𝜆𝑑𝑙 𝜀0 = 𝜆𝑙 𝜀0 𝐸 = 𝜆𝑙 2𝜋𝑟𝑙𝜀0 = 𝜆 2𝜋𝑟𝜀0 𝐸 𝑟 = 𝑅 2 = 𝜆 2𝜋𝑅 2 𝜀0 = 𝜆 𝜋𝑅𝜀0 𝐹 = 𝑞𝐸 = 𝑚𝑎 𝑎 = 𝑞𝐸 𝑚 = 𝑞 𝑚 𝜆 𝜋𝑅𝜀0 𝒂 = 𝒒𝝀 𝝅𝑹𝒎𝜺𝟎