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The Electric Field II: Continuous Charge Distributions

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The Electric Field II: Continuous Charge Distributions TIPL22_727-762-hr 3/30/07 8:41 AM Page 727 CHAPTER22 LIGHTNING IS AN ELECTRIC PHENOMENA. DURING A LIGHTNING The Electric Field II: STRIKE, CHARGES ARE TRANSFERRED BETWEEN THE CLOUDS AND THE Continuous Charge GROUND. THE VISIBLE LIGHT GIVEN OFF COMES FROM AIR MOLECULES Distributions RETURNING TO LOWER ENERGY STATES. (Photo Disc.) S 22.1 CalculatingE from Coulomb’s Law How would you calculate the 22.2 Gauss’s Law ? charge on the surface of Earth? S (See Example 22-15.) 22.3 Using Symmetry to CalculateE with Gauss’s Law 22.4 Discontinuity of En 22.5 Charge and Field at Conductor Surfaces * 22.6 The Equivalence of Gauss’s Law and Coulomb’s Law in Electrostatics n a microscopic scale, charge is quantized. However, there are often situa- tions in which many charges are so close together that the charge can be thought of as continuously distributed. We apply the concept of density to charge similarly to the way we use it to describe matter. In addition to continuous charge distributions, we examine the impor- O tance of symmetry within the electric field. The mathematical findings of Carl Friedrich Gauss show that every electric field maintains symmetric proper- ties. It is an understanding of charge distribution and symmetry within the electric field that aids scientists in a vast array of fields. In this chapter, we show how Coulomb’s law is used to calculate the electric field produced by various types of continuous charge distributions. We then introduce Gauss’s law and use it to calculate the electric fields produced by charge distributions that have certain symmetries. 727 TIPL22_727-762-hr 3/29/07 10:31 AM Page 728 728 | CHAPTER 22 The Electric Field II: Continuous Charge Distributions S 22-1 CALCULATING E FROM COULOMB’S LAW Figure 22-1 shows an element of charge dq ϭ r dV that is small enough to be con- dq = ρ dV P sidered a point charge. The element of charge dq is the amount of charge in volume k dq r dE = r^ r 2 element dVSand is the charge per unit volume. Coulomb’s law states that the elec- r tric field dE at a field point P due to this element of charge is S k dq dE ϭ dE rn ϭ rn 22-1a r r2 n where r is a unit vector directed away from the charge element dq and toward FIGURE 22-1 S S An element of charge dq point P, and dE (the component of dE in the direction of rn ) is given by k dq r2. produces a field dE ϭ (k dq r2)rn at point r S > > The total field E at P is calculated by integrating this expression over the entire P. The field at P is calculated by integrating charge distribution. That is, Equation 22-1a over the entire charge distribution. S S krn E ϭ Ύ dE ϭ Ύ dq 22-1b r2 ELECTRIC FIELD DUE TO A CONTINUOUS CHARGE DISTRIBUTION The use of a continuous charge density to describe a large number of discrete charges is similar to the use of a continuous mass density to describe air, which ac- tually consists of a large number of discrete atoms and molecules. In both cases, it is usually easy to find a volume element ¢V that is large enough to contain a mul- titude of individual charge carriers and yet is small enough that replacing ¢V with a differential dV and using calculus introduces negligible error. If the charge is dis- The x component of rn is dq ϭ s dA dq ϭ l dL n # n tributed over a surface or along a line, we use or and inte- ! r i ϭ cos u, where u is the angle n grate over the surface or line. (In these cases and ls are the charge per unit area between rn and i .* The and zy and charge per unit length, respectively.) The integration usually is done by ex- components of rn are calculated in n pressing r in terms of its Cartesian components, and then integrating one compo- like manner. nent at a time. PROBLEM-SOLVING STRATEGY S Calculating E Using Equations 22-1a and 22-1b PICTURE SketchS the charge configuration along with a field point P (the point where E is to be calculated). In addition, the sketch should include an increment of charge dq at an arbitrary source point S. SOLVE 1. Add coordinate axes to the sketch. The choice of axes should exploit any symmetry of the charge configuration. For example, if the charge is along a straight line, then select that line as one of the coordinate axes. Draw a second axis that passes through the field point P. In addition, include the coordinates of both P and S, the distance r between P and S, and the unit vector rn directed away from S toward P. S S 2. To compute the electric field E using Equation 22-1b, we express dE ϭ dE rn S n # n r in component form. The x component of dE is dE ϭ dE r i ϭ dE cos u, n x r r where u is the angle between rn and i (see Figure 22-2), and the y component S n # n ϭ ϭ u of dE is dEy dEr r j dEr sin . * The componet of a vector in a given direction is equal to the scalar product of the vector with the unit vector in the given direction. Scalar products are discussed in Section 6-3. TIPL22_727-762-hr1 10/10/07 3:01 PM Page 729 S Calculating E from Coulomb’s Law SECTION 22-1 | 729 S 3. Express E in Equation 22-1b in terms of its and yx components: k dq E ϭ Ύ dE ϭ Ύ dE cos u ϭ Ύ cos u x x r r2 k dq E ϭ Ύ dE ϭ Ύ dE sin u ϭ Ύ sin u y y r r2 r s l 4. To calculate Ex , express dq as dV or dA or dL (whichever is appropriate) and integrate. To calculate Ey , follow a procedure similar to that used for calculating Ex . 5. Symmetry argumentsS are sometimes used to show that one or more components of E are equal to zero. (For example, a symmetry argument ϭ is used to show Ey 0 in Example 22-5.) CHECK If the charge distribution is confined to a finite region of space at See points far from the charge distribution, the expression for the electric field Math Tutorial for more will approach that of a point charge located at the center of charge. (If the information on charge configuration is sufficiently symmetric then the location of the center of charge can be obtained by inspection.) Trigonometry Example 22-1 Electric Field Due to a Line Charge of Finite Length A thin rod of length L and charge Q is uniformly charged, so it has a linear charge density l ϭ Q L. Find the electric field at point P, where P is an arbitrarily positioned > point. PICTURE Choose the x axis so the rod is on the x axis between points x and x , and 1 2 y choose the y axis to be through the field pointS P. Let yP be the radial distance of P from the x axis. To calculate the electric field E at P, Swe separately calculate Ex and Ey . Using Equations 22-1, first find the field increment E at Pd due to an arbitrary increment dE S P dq of the charge distribution. Then integrate each component of dE over the entire charge distribution. (Because Q is distributed uniformly, the linear charge density l equals Q L.) > Q λ = SOLVE L ˆ − j 1. Sketch the charge configuration and the field point P. Include the and yx axes with L = x2 x1 r1 λ yP thex axis lying along the line of charge and the y axis passing through P. In addition, dq = dxs rˆ ϭ r sketch an arbitrary increment of the line charge at point S (at x xS) that has a r2 lengthdxS andS a charge dq, and the electric field at P due to dq. Sketch the electric Q ˆ field vector dE as if dq is positive (Figure 22-2): θ S θ i S S 1 θ ϭ n ϩ n ϭ n 2 2. E E i E j. Find expressions for dE and dE in dE dE r +++++++++ x y x y r x u x1 x2 termsS of dEr and , where dEr is the component of # n ϭ n ϭ u dq − so dEx dEr r i dEr cos xs = xs dE in the direction away from toward PS : dx # n s 0 ϭ n ϭ u dEy dEr r j dEr sin Ϫ FIGURE 22-2 Geometry for the k dq xS 3. First we solve for E . Express dE using Equation dE ϭ and cos u ϭ calculation of the electric field at field x r r r2 r 21-1a, where r is the distance from the source pointP due to a uniformly charged rod. so point S to the field point P. We see (Figure 22-2) ϭ ϭϪ that cos u ƒx ƒ r x r. In addition, use u l S > S> k dq k cos dxS ϭ dE ϭ cos u ϭ l dq dxS: x r2 r2 x x u l u 2 k cos dxS 2 cos dxS 4.
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