Chapter 24 electric flux 24 1 Electric Flux

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 1

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

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 3

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 angle θ with the perpendicular to the surface • Then ΦE = EA cos θ • The flux is a maximum when the surface is perpendicular to the field • The flux is a minimum (zero) when the surface is (parallel) to the field Norah Ali Al-moneef King Saud university 4

Electric Flux • If the field varies over the surface, Φ = EA cos θ 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 5

Electric Flux • The surface integral means the integral must be evaluated over the surface in question • The value of the flux depends both on the field pattern and on the surface • SI units: N. m 2/C 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 < 90 o, Φ > 0 • (2) θ = 90 o, Φ = 0 • (3) 180 o > θ > 90 o, 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 field lines pass through two surfaces perpendicularly and are parallel to the other four surfaces For side 1, ΦE = -El 2 For side 2, ΦE = El 2 For the other sides, ΦE = 0 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/r 2 Gauss’ Law Norah Ali Al-moneef King Saud university The flux of electric field through a closed surface is proportional to the charge enclosed. 9 9

• Gaussian surfaces of various shapes can surround the charge (only S 1 is spherical) • The electric flux is proportional to the number of electric field lines penetrating these surfaces, and this number is the same • Thus the net flux through any closed surface surrounding a point charge q is given by q/εo and is independent of the shape of the surface Norah Ali Al-moneef King Saud university 10

Gauss’ Law • If the charge is outside the closed surface of an arbitrary shape, then any field line entering the surface leaves at another point • Thus the electric flux through a closed surface that surrounds no charge is zero Norah Ali Al-moneef King Saud university 11

Gauss’ Law • Since the electric field due to many charges is the vector sum of the electric fields produced by the individual charges, the flux through any closed surface can be expressed as • Although Gauss’s law can, in theory, be solved to find for any charge configuration, in practice it is limited to symmetric situations • One should choose a Gaussian surface over which Ali Al-moneef the surface integral. Norah can be simplified and the 12 King Saud university

Norah Ali Al-moneef King Saud university 13

Problem 24 -9 The following charges are located inside a submarine: 5. 00 μC, – 9. 00 μC, 27. 0 μC, and – 84. 0 μC. (a) Calculate the net electric flux through the hull of the submarine. (b) 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 14

Problem solving guide for Gauss’ law Use the symmetry of the charge distribution to determine the pattern of the field lines. Choose a Gaussian surface for which E is either parallel to or perpendicular to d. A. If E is parallel to d. A, then the magnitude of E should be constant over this part of the surface. The integral then reduces to a sum over area elements. Norah Ali Al-moneef King Saud university 15

Gauss’ Law and Coulomb law Example 24. 4 The Electric Field Due to a Point Starting with Gauss’s law, calculate the electric field due to an Charge isolated point charge q. The field lines are directed radially outwards and are perpendicular to the surface at every The angle point, so between E and d. A 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 Integral is the sum of surface area Coulomb’s Law Norah Ali Al-moneef King Saud university 16

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. (A)Find the magnitude of the E-field at a point outside the sphere (B)Find • Forthe magnitude r > aof the E-field at a point inside the sphere Norah Ali Al-moneef King Saud university 17

b--Find the magnitude of the E-field at a point inside the sphere r< a r Now we select a spherical Gaussian surface with radius r < a. Again the symmetry of the charge distribution allows us to simply evaluate the left side of Gauss’s law just as before. a Q The charge inside the Gaussian sphere is no longer Q. If we call the Gaussian sphere volume V’ then“V Volume charge density“: ρ = charge / unit volume is used to characterize the charge distribution Norah Ali Al-moneef King Saud university 18

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 19

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 20

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 21

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 22

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. Symmetry Þ 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 (the direction of a surface is its normal!) at every point on the curved part of the surface (the body of the cylinder. Norah Ali Al-moneef King Saud university 23

Line of Charge Norah Ali Al-moneef King Saud university 24

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. Assume that we have a thin, infinite nonconducting sheet of positive charge The charge density in this case is the charge per unit area, 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 25

To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height 2 r, chosen to cut through the plane perpendicularly. Because the electric field is perpendicular to the plane everywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the ends of the cylinder. Using Gauss’ Law we get the electric field from an infinite non-conducting sheet with charge density Norah Ali Al-moneef King Saud university 26

24 -4 Conductors 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 27

24 -4 Conductors 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. If there was a field in the conductor the charges would accelerate under the action of the field. ++++++ ----------- The charges in the conductor move creating an internal electric field that cancels the applied field on the inside of the conductor Ein E E Norah Ali Al-moneef King Saud university 28

When electric charges are at rest, the electric field within a conductor is zero. On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest. The electric field is stronger where the surface is more sharply curved. Norah Ali Al-moneef King Saud university 29

• Charge Resides on the Surface • Choose a Gaussian surface inside but close to the actual surface • The electric field inside is zero • There is no net flux through the gaussian surface • Because the gaussian surface can be as close to the actual surface as desired, there can be no charge inside the surface Since no net charge can be inside the surface, any net charge must reside on the surface Gauss’s law does not indicate the distribution of these charges, only that it must be on the surface of the conductor Norah Ali Al-moneef King Saud university 30

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 31

Under electrostatic conditions the electric field inside a solid conducting sphere is zero. Outside the sphere the electric field drops off as 1 / r 2, as though all the excess charge on the sphere were concentrated at its center. Electric field = zero (electrostatic) inside a solid conducting sphere Norah Ali Al-moneef King Saud university 32

homework problem 1 - Charge in a Cube Q=3. 76 n. C is at the center of a cube. Q 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 (i) To the left of the plates? (ii) To the right of the plates? (iii) Between the plates? Norah Ali Al-moneef King Saud university 33

problem 3 : 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 problem 4 : A 5 n. C 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 34

problem 5 : 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. 0 x 103 N/C? problem 6 : 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 (a)10. 0 cm and (b)20. 0 cm from the center of the charge distribution. Norah Ali Al-moneef King Saud university 35

problem 7 : A sphere of radius 8. 0 cm carries a uniform volume charge density r = 500 n. C/m 3. 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 Norah Ali Al-moneef field outside the cylinder, a distance r from the axis. 36 King Saud university

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 k. N/C directed perpendicularly to the plate. Find (a) the charge density of each face of the plate and (b) the total charge on each face. Norah Ali Al-moneef King Saud university 37

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 k. N/C directed perpendicularly to the plate. Find (a) the charge density of each face of the plate and (b) the total charge on each face. Norah Ali Al-moneef King Saud university 38

Summary: Gauss’ Law · Gauss’ Law depends on the enclosed charge only 1. 2. 3. 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 39

GAUSS LAW – SPECIAL SYMMETRIES SPHERICAL CYLINDRICAL PLANAR (point or sphere) (line or cylinder) (plane or sheet) CHARGE DENSITY from central point Depends only on perpendicular distance from line Depends only on perpendicular distance from plane Sphere centered at point of symmetry Cylinder centered at axis of symmetry Pillbox or cylinder with axis perpendicular to plane E constant at surface E ║A - cos = 1 E constant at curved surface and E ║ A E ┴ A at end surface cos = 0 E constant at end surfaces and E ║ A E ┴ A at curved surface cos = 0 Depends only on radial distance GAUSSIAN SURFACE ELECTRIC FIELD E Norah Ali Al-moneef King Saud university 40

Geometry Charge Density Gaussian surface Linear l = q/L Cylindrical, with axis along line of charge Sheet or Plane s = q/A Cylindrical, with axis along E. Spherical r = q/V Spherical, with center on center of sphere Electric field Norah Ali Al-moneef King Saud university Line of Charge Conducting Nonconducting r R r < R 41

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 n. C (b) 120 p. C (c) 36 m. C (d) 12 C (e) 3. 0 n. C 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 11/22/2020 Norah Ali Al-moneef king saud university 42

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, 8 x 10 -9 C and -2 x 10 -9 C are separated by 4 m. The electric field magnitude (in units of V/m) midway between them is: A)9 x 109 B) 13, 500 C) 135, 000 D) 36 x 10 -9 11/22/2020 E) 22. 5 Norah Ali Al-moneef king saud university 43