Electric Flux. To investigate this, we have to understand electric flux.

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1 Problem A charge q 1 = +5. nc is placed at the origin of an xy-coordinate system, and a charge q 2 = -2. nc is placed on the positive x-axis at x = 4. cm. (a) If a third charge q 3 = +6. nc is now placed at the point x = 4. cm, y = 3. cm. find the x, y components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force. PHYS 153 8W 1

2 Problem 21.8 Three point charges are placed on the y-axis: a charge q at y = a, a charge -2q at the origin, and a charge q at y = -a. Such an arrangement is called an electric quadrupole. (a) Find the magnitude and direction of the electric field at points on the positive x-axis. (b) Use the binomial expansion for the electric field valid for x >> a. Contrast this to that of the electric field of a point charge and that of the electric field of a dipole. PHYS 153 8W 2

3 lectric Flux We have examined the question, given the charge distribution, what is the electric field at point P? Is it possible to study the reverse of this? Given an electric field pattern for a given region, what is the charge distribution in that region? To investigate this, we have to understand electric flux. field of a point charge field at the surface of a sphere containing a single point charge. The electric field points out of the sphere. PHYS 153 8W 3

4 field of two charges field at the surface of a sphere containing two point charges. field of a dipole field at the surface of a sphere containing a dipole PHYS 153 8W 4

5 field at the surface of a sphere containing no charge. Notice the difference among these figures. Two positive charges have twice the number of field lines at the surface than one charge. For a dipole inside a sphere, the net number of lines leaving and entering the surface is zero. If there is no charge at the surface, there are no lines leaving or entering the surface. The mathematical quantity related to the number of field lines crossing the surface is called the electric flux. The electric flux Φ through a surface of area A that is perpendicular to the field is defined as Φ =A PHYS 153 8W 5

6 Concept of the idea area perpendicular to the electric field. In (a), the area is perpendicular to the field and the maximum number of lines pass through. In (b), the area is tilted to the field, and fewer lines pass through it. In (c), the area is parallel to the field, and no lines pass through It. Area When = Acosφ φ= =9 When, Φ = A o φ Φ = Fig (maximum) (minimum) So in general, Φ =Acosφ PHYS 153 8W 6

7 Φ = A= A or A Φ =. A How do we identify the direction of the area A? n For a closed surface, the area vector points outward. A= An For a single flat surface as in Fig. 22.6, the area direction must be specified. For a non uniform field (or surface area) Φ =. d A Refer to Fig again. The flux through the area in (a) is a maximum. The flux through the area in (c) is zero. Gauss s Law Refer back to slide 3. Suppose the radius of the sphere is r. PHYS 153 8W 7

8 1 4πε At the surface of the sphere which encloses charge q, 2 Φ A 1 πr 2 =. =.4 = 2 4πε r q q ε This is a statement of Gauss s law which states that the total electric flux through any closed surface equals the total electric charge inside the surface divided by ε. = q r Φ =. da= Q encl ε Gauss s law The circle on the integral sign indicates integration over a closed surface. Try to think why this is independent of radius for a sphere. The surface enclosing the charge can be any shape. PHYS 153 8W 8

9 e.g. non-spherical surface. Figure An irregular shaped surface encloses a sphere which encloses a charge +q. The flux line is to any da on the sphere, but is not to any da on the irregular surface. For the irregular surface, the flux Φ = da=( cosφ) da For a spherical surface at the same distance from q, the flux Φ =( dacosφ) since the projection of da onto the spherical surface is dacosφ Hence the results for the irregular shaped surface, and the spherical surface are the same. PHYS 153 8W 9

10 Applications (a) Field of a charged conducting sphere with charge q. Utilize the symmetry. The field must be radially outward, as shown, and depend only on the distance r (spherical symmetry). Hence choose as a Gaussian surface a sphere or radius r concentric with the charged sphere. Fig This is the same as for a point charge q at the centre. 2. da=.4πr = = 1 4πε q ε q 2 r PHYS 153 8W 1

11 d. A = Q ε Inside the sphere, since all of the charge is on the = = Surface. Hence inside a conductor. The plot of vs. r in Fig shows how varies with r from r= to r>r. (b) inside and outside a uniformly charged solid (non-conducting) sphere Radius R and total charge Q uniformly distributed. Charge density Q ρ = = V Q πr Again use spherical symmetry. Using a Gaussian sphere of radius r>r, Gauss s law gives r 1 Q = 2 ( r R 4πε r ) For r<r, is not zero. At a given r, the charge enclosed is PHYS 153 8W 11

12 4 Q 4 = ρ π π 3 πr Q inside. r = ( )( r ) = 4 3 r Q R For the electric field inside the sphere, Gauss s law then gives r 4 2 π r = Q ε r R 1 Q r = r 3 ( r R) 4πε R (c) Field of a line charge Fig (Tipler) What is the electric field due to a very long, uniformly charged, wire? PHYS 153 8W 12

13 Symmetry arguments suggest that the electric field lines are radial and lie in planes perpendicular to the wire. Fig Choose as a Gaussian surface a cylinder with the wire lying along its axis. λ is the charge per unit length. For a given length l, = 1 2πε λ r. da= (2πrl) = λl ε (recall the same result derived previously) (d) Field of an infinite plane sheet of charge. σ is the charge per unit area on a thin, flat, infinitely large sheet. Planar symmetry suggests that is perpendicular everywhere to the sheet. PHYS 153 8W 13

14 Choose as our Gaussian surface a cylinder with its axis perpendicular to the sheet. area A From Gauss s law, 2A= σa ε = σ 2ε This result was also found previously. Imagine now the plane becoming a conductor of non zero thickness. The same geometry above applies, but with the bottom of the cylinder inside the conductor. Since = inside the conductor, protrudes only from one end of the cylinder. Hence, at the surface of a conductor, σ = PHYS 153 8W ε 14

15 Faraday cage Sensitive electronic equipment in an environment where there is a lot of electrical interference can be protected by means of a Faraday cage. e.g. in a radio astronomy observatory, very sensitive radio receivers operate next to computers which are often very electrically noisy. Inside the conducting box, = (Gauss s law). This conducting box is a Faraday cage, and can shield sensitive equipment from stray electrostatic fields, or electromagnetic radiation. Fig PHYS 153 8W 15

16 Problem A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its centre. The total charge on the shell is -3Q, and it is insulated from its surroundings. (a) Derive expressions for the electric field magnitude in terms of distance r from the centre for the regions r<a, a<r<b, r>b. (b) What is the surface charge density on the inner surface of the conducting shell? (c) What is the surface charge density on the outer surface of the conducting shell? (d) Sketch the electric field lines and the locations of all charges. (e) Make a plot of the electric field magnitude as a function of r.