Chapter 24 electric flux

1. Chapter 24 electric flux 24.1 Electric Flux 24.2 Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions • A Spherically Symmetric Charge Distribution (An insulating solid sphere ) • The Electric Field Due to a Point Charge • The Electric Field Due to a Thin Spherical Shell • The Electric Field Due to A Cylindrically Symmetric Charge Distribution • the electric field due to an infinite plane • 24.4 Conductors in Electrostatic Equilibrium Norah Ali Al-moneef King Saud university

2. We will use a “charge density” to describe the distribution of charge. This charge density will be different depending on the geometry Norah Ali Almoneef 2 Norah Ali Al-moneef King Saud university

3. Electric Flux • Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the field ΦE = EA Norah Ali Al-moneef King Saud university

4. 24.1 Electric Flux • The electric flux is proportional to the number of electric field lines penetrating some surface • The field lines may make some angleθwith the perpendicular to the surface • Then ΦE = EAcosθ Norah Ali Al-moneef King Saud university

5. Electric Flux • If the field varies over the surface, Φ = EAcosθ is valid for only a small element of the area • In the more general case, look at a small area element • In general, this becomes Norah Ali Al-moneef King Saud university

6. Electric Flux, Closed Surface • For a closed surface, by convention, the A vectors are perpendicular to the surface at each point and point outward • (1) θ < 90o, Φ > 0 • (2) θ = 90o, Φ = 0 • (3) 180o > θ > 90o, Φ < 0 • The net flux through the surface is proportional to the number of lines leaving the surface minus the number entering the surface Norah Ali Al-moneef King Saud university

7. Example 24-1: flux through a cube of • a uniform electric field • The flux through four of the faces 3, 4 and unnamed ones is ΦE = 0 because E is perpendicular to dA onthese faces • For side 1, ΦE = -El 2 • For side 2, ΦE = El 2 Therefore, Φtotal = 0 Norah Ali Al-moneef King Saud university

8. 24-2 Electric lines of flux and Derivation of Gauss’ Law using Coulombs law • Consider a sphere drawn around a positive point charge. Evaluate the net flux through the closed surface. Net Flux = For a Point charge E=kq/r2 Gauss’ Law The flux of electric field through a closed surface is proportional to the charge enclosed. Gauss’ Law 8 Norah Ali Al-moneef King Saud university

9. Gauss’ Law Norah Ali Al-moneef King Saud university

10. Problem 24-9 The following charges are located inside a submarine: 5.00 μC, –9.00 μC, 27.0 μC, and –84.0 μC. • Calculate the net electric flux through the hull of the submarine. • Is the number of electric field lines leaving the submarine greater than, equal to, or less than the number entering it? (b) Since the net electric flux is negative, more lines enter than leave the surface. Norah Ali Al-moneef King Saud university

11. Gauss’ Law and Coulomb law Example 24.4 The Electric Field Due to a Point Charge Starting with Gauss’s law, calculate the electric field due to an isolated point charge q. • The field lines are directed radially outwards and are perpendicular to the surface at every point, so The angle  between E and dA is zero at any point on the surface, we can re-write Gauss’ Law as E has the has same value at all points on the surface E is can be moved out A spherical Gaussian surface centered on a point charge q Coulomb’s Law Norah Ali Al-moneef King Saud university

12. Example 24.5 A Spherically Symmetric Charge Distribution An insulating solid sphere of radius a has a uniform volume charge density ρ and carries total charge Q. • Find the magnitude of the E-field at a point outside the sphere • Find the magnitude of the E-field at a point inside the sphere • For r > a Norah Ali Al-moneef King Saud university

13. r b--Find the magnitude of the E-field at a point inside the sphere r < a Let the Volume of the smaller sphere where r < a is V’. So Therefore Gauss’s law in the region r < a is a Q Norah Ali Al-moneef King Saud university

14. Field Due to a Spherically Symmetric Charge Distribution • Inside the sphere, E varies linearly with r (E → 0 as r → 0) • The field outside the sphere is equivalent to that of a point charge located at the center of the sphere Norah Ali Al-moneef King Saud university

15. Example 24.6 The Electric Field Due to a Thin Spherical Shell A thin spherical shell of radius a has a total charge Q distributed uniformly over its surface Find the electric field at points (A) outside and (B) inside the shell. Norah Ali Al-moneef King Saud university

16. Let’s start with the Gaussian surface outside the sphere of charge, r > a • We know from symmetry arguments that the electric field will be radial outside the charged sphere • If we rotate the sphere, the electric field cannot change • Spherical symmetry • Thus we can apply Gauss’ Law and get • … so the electric field is Norah Ali Al-moneef King Saud university

17. Let’s let’s take the Gaussian surface inside the sphere of charge, r < a • We know that the enclosed charge is zero so • We find that the electric field is zero everywhere inside spherical shell of charge • Thus we obtain two results • The electric field outside a spherical shell of charge is the same as that of a point charge. • The electric field inside a spherical shell of charge is zero. Norah Ali Al-moneef King Saud university

18. 24-6 The Electric Field Due to A Cylindrically Symmetric Charge Distribution Find the E-field a distance r from a line of positive charge of infinite length and constant charge per unit length λ. • SymmetryÞ E field must be ^ to line and can only depend on distance from line • Select a cylinder as Gaussian surface. The cylinder has a radius of r and a length of ℓ • is constant in magnitude and parallel to the surface Norah Ali Al-moneef King Saud university

19. Line of Charge Norah Ali Al-moneef King Saud university

20. Example 24.8 A Plane of Charge • Find the electric field due to an infinite plane of positive charge with uniform surface charge density σ. • The charge density in this case is the charge per unit area, i.e =q/A • From symmetry, we can see that the electric field will be perpendicular to the surface of the sheet Norah Ali Al-moneef King Saud university

21. To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a cylinder with cross sectional area A • Using Gauss’ Law we get • the electric field from an infinite plane positive sheet with charge density  is Norah Ali Al-moneef King Saud university

22. 24-4Conductors in Electrostatic Equilibrium By electrostatic equilibrium we mean a situation where there is no net motion of charge within the conductor • The electric field is zero everywhere inside the conductor • Any net charge resides on the conductor’s surface • The electric field just outside a charged conductor is perpendicular to the conductor’s surface • On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest. Norah Ali Al-moneef King Saud university

23. 24-4Conductors in Electrostatic Equilibrium • The electric field is zero everywhere inside the conductor Why is this so? • Place a conducting slab in an external field, E. Norah Ali Al-moneef King Saud university

24. E-Field’s Magnitude and Direction Choose a cylinder as the Gaussian surface The field must be perpendicular to the surface If there were a parallel component to E, charges would experience a force and accelerate along the surface and it would not be in equilibrium • The net flux through the surface is through only the flat face outside the conductor • The field here is perpendicular to the surface • Applying Gauss’s law Norah Ali Al-moneef King Saud university

25. Under electrostatic conditions the electric field inside a solid conducting sphere is zero. Outside the sphere the electric field drops off as 1 / r2, as though all the excess charge on the sphere were concentrated at its center. Norah Ali Al-moneef King Saud university

26. Q homework problem1 - Charge in a Cube • Q=3.76 nC is at the center of a cube. What is the electric flux through one of the sides? Problem 2 : Two large thin metal plates with surface charge densities of opposite signs but equal magnitude of 44.27 ×10–20 Cm–2 are placed parallel and close to each other. What is the field • To the left of the plates? (ii) To the right of the plates? (iii) Between the plates? Norah Ali Al-moneef King Saud university

27. problem3 : A spherical shell of radius 10 cm has a charge 2×10–6 C distributed uniformly over its surface. Find the electric field (a) Inside the shell (b) Just outside the shell (c) At a point 15 cm away from the centre problem4 : A 5 nC charge is placed inside of a cubic surface. The cube is measured to be 0.5 m per side. What is the flux through the cube? Norah Ali Al-moneef King Saud university

28. problem5 : Ex) What is the electric flux through this disk of radius r = 0.10 m if the uniform electric field has a magnitude E = 2.0x103 N/C? problem6 : Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field • 10.0 cm and • 20.0 cm from the center of the charge distribution. Norah Ali Al-moneef King Saud university

29. problem7 : A sphere of radius 8.0 cm carries a uniform volume charge density r = 500 nC/m3. What is the electric field magnitude at r = 8.1 cm? Problem 8: A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of λ, and the cylinder has a net charge per unit length of 2λ. From this information, use Gauss’s law to find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b) the electric field outside the cylinder, a distance r from the axis. Norah Ali Al-moneef King Saud university

30. Problem 9: A uniformly charged, straight filament 7.00 m in length has a total positive charge of 2.00 μC. An uncharged cardboard cylinder 2.00 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder and (b) the total electric flux through the cylinder. Problem 10: A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find • the charge density of each face of the plate and • the total charge on each face. Norah Ali Al-moneef King Saud university

31. Problem 11: A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find • the charge density of each face of the plate and • the total charge on each face. Norah Ali Al-moneef King Saud university

32. Summary: Gauss’ Law ·Gauss’ Law depends on the enclosed charge only • If there is a positive net flux there is a net positive charge enclosed • If there is a negative net flux there is a net negative charge enclosed • If there is a zero net flux there is no net charge enclosed • Gauss’ Law works in cases of symmetry Norah Ali Al-moneef King Saud university

33. GAUSS LAW – SPECIAL SYMMETRIES SPHERICAL (point or sphere) CYLINDRICAL (line or cylinder) PLANAR (plane or sheet) Depends only on radial distance CHARGE DENSITY Depends only on perpendicular distance from line Depends only on perpendicular distance from plane from central point ELECTRIC FIELD E GAUSSIAN SURFACE Sphere centered at point of symmetry Cylinder centered at axis of symmetry Pillbox or cylinder with axis perpendicular to plane E constant at end surfaces and E ║ A E ┴ A at curved surface cos  = 0 E constant at curved surface and E ║ A E ┴ A at end surface cos  = 0 E constant at surface E ║A - cos  = 1 Norah Ali Al-moneef King Saud university

34. Line of Charge Conducting Nonconducting r  R r < R Norah Ali Al-moneef King Saud university

35. 1- The electric field has a magnitude of 3.0 N/m at a distance of 60 cm from a point charge. What is the charge? (a) 1.4 nC (b) 120 pC (c) 36 mC (d) 12 C (e) 3.0 nC 2- An electron traveling horizontally from North to South enters a region where a uniform electric field is directed downward. What is the direction of the electric force exerted on the electron once it has entered the field? (a) downward (b) upward (c) to the east (d) to the west (e) to the north Norah Ali Al-moneef king saud university

36. A conducting sphere has a net charge of −4.8 × 10−17 C. What is the approximate number of excess electrons on the sphere? (a) 100 (b) 200 (c) 300 (d) 400 (e) 500 Two point charges, 8x10-9 C and -2x10-9 C are separated by 4 m. The electric field magnitude (in units of V/m) midway between them is: • 9x109 B) 13,500 C) 135,000 D) 36x10-9 E) 22.5 Norah Ali Al-moneef king saud university