Chapter 2 Electrostatics Action at a distance 2.1 The Electric Field: 2.1.1 Introduction What is the force on the test charge Q due to a source Coulomb’s law, like Newton’s law of gravitation, involves the charge q? concept of action at a distance. We shall consider the special case of the electrostatics in It simply states how the particles interact but provides no which all the source charges are stationary. explanation of the mechanism by which the force is The principle of superposition states that the interaction transmitted from one point to the other. between any two charges is completely unaffected by the Even Newton himself is not comfortable with this aspect of presence of others. his theory. What is the concept of action at a distance? This leads to the gravitational, electric, and magnetic field. 1 2 2.1.2 Coulomb’s Law 2.1.3 The Electric Field How does one particle sense the presence of the other? Coulomb’s law quantitatively describe the interaction of charges. The electric charge creates an electric field in the space around it. A second charged particle does not interact Coulomb determined the force law for electrostatic charges directly with the first; rather, it responds to whatever field it directly by experiment. encounters. In this sense, the field acts as an intermediary kqQ 1 qQ between the particles. F = 2 rˆ = 2 rˆ Q q1 q2 r 4πε0 r F = F1 + F2 +" = ( 2 rˆ1 + 2 rˆ2 +") = QE 2 4πε 0 r1 r2 9 N ⋅m n Where k = 9×10 2 1 q q 1 q C 1 2 " i where E = ( 2 rˆ1 + 2 rˆ2 + ) = ∑ 2 rˆi 2 4πε 0 r1 r2 4πε 0 i=1 ri −12 C and ε 0 = 8.86×10 N ⋅m2 The electric field strength is defined as the force per 3 unit charge placed at that point. 4 Example 2.1.4 Continuous Charge Distributions On a clear day there is an electric field of approximately In order to find the electric field due to a continuous distribution 100 N/C directed vertically down at the earth’s surface. of charge, one must divide the charge distribution into Compare the electrical and gravitational forces on an infinitesimal elements of charge dq which may be considered to electron. be point charges. 1 dq 1 dq dE = ˆ ⇒ E = ˆ Solution: 2 r ∫ 2 r 4πε 0 r 4πε 0 r The magnitude of the electrical force is Fe=eE=1.6x10-19x100=1.6x10-17 N. (upward) The magnitude of the gravitational force is Fg=mg=9.11x10-31x9.8=8.9x10-30 N. (downward) 5 6 Example 2.1 Example Non-conducting disk of radius a has a uniform surface What is the field strength at a distance R from an infinite charge density σ C/m2. What is the field strength at a line of charge with linear charge density λ C/m. distance y from the center along the central axis. Solution: Solution: Since the charge carrier is infinite long, the electric field in y-direction completely The y-component of the field is cancel out. Thus the resultant field is kdq y along the x-axis. dEy == dE cosθ rr2 222 1 λ cosθdA 1 λRsec2 θ cosθ dθ where rdqxdx=+x y and =σπ(2 ) dE = = x 4πε r 2 4πε (Rsecθ )2 a 2xdx 0 0 Eky= πσ y ∫0 223/2 1 λ cosθdθ ()xy+ = 2 4πε R a dx 0 = πσky ∫0 223/2 1 λ π / 2 λ ()xy+ Ex = cosθdθ = ∫−π / 2 4πε0 R 2πε0 R 7 8 2.2 Divergence and Curl of Electrostatics Fields Field Lines 2.2.1 Field Lines, Flux, and Gauss’s Law How to determine the field strength from the field lines? How to express the magnitude and vector properties of the field strength? The lines are crowed together when the field is strong and spread apart where the field is weaker. The field strength is The field strength at any point could be represented by an proportional to the density of the lines. arrow drawn to scale. However, when several charges are present, the use of arrows of varying length and orientations becomes confusing. Instead we represent the electric field by continuous field lines or lines of force. 9 10 Example Flux Sketch the field lines for two point charges 2Q and –Q. The electric flux ΦE through this surface is defined as Solution: Φ E = EAcosθ = E⋅ A (a)Symmetry (b)Near field (c)Far field For a nonuniform electric field (d)Null point (e)Number of lines Φ = E⋅nˆda E ∫ 11 12 Flux Gauss’s Law Flux leaving a closed surface is positive, whereas flux How much is the flux for a spherical Gaussian surface around entering a closed surface is negative. a point charge? The net flux through the surface is zero if the number of The total flux through this closed lines that enter the surface is equal to the number that Gaussian surface is leave. Φ = E⋅nˆda E ∫ Q 2 Q = 2 ⋅4π r = 4πε0r ε 0 The net flux through a closed surface equals 1/ε0 times the net charge enclosed by the surface. Can we prove the above statement for arbitrary closed shape? 13 14 Gauss’s Law (II) Turn Gauss’s Law from integral equation into differential form •The circle on the integral sign indicates that the Gaussian Q surface must be enclosed. E⋅da = E⋅nˆda = enc ∫S ∫S •The flux through a surface is determined by the net charge ε 0 enclosed. By applying the divergence theorem Q 1 E⋅nˆda = (∇ ⋅E)dτ and enc = ρ dτ ∫S ∫v ∫v ε 0 ε 0 1 So (∇ ⋅E)dτ = ρdτ ∫v ∫v ε 0 Since this holds for any volume, the integrands must be equal How to apply Gauss’s law? 1 ∇ ⋅E = ρ Gauss’s Law in differential form. 1. Use symmetry. ε 0 2. Properly choose a Gaussian surface (E//A or E⊥A). 15 16 2.2.2 The Divergence of Electric Field & Example 2.2 2.2.3 Application of the Gauss’s Law The electric field can be expressed in the following form A non-conducting uniform charged sphere of radius R has a total charge Q uniformly distributed throughout its volume. 1 dq 1 ˆ E = ˆ = r ρ(r′)dτ ′ Find the field (a) inside, and (b) outside, the sphere. ∫all space 2 r ∫all space 2 4πε 0 r 4πε 0 r Solution: Divergence of the electric field is (a) inside Why the divergence 4 3 1 ˆ Φenc Q 3 πr 1 r ′ ′ operator doesn’t apply E = rˆ = ( ) rˆ ∇ ⋅E = (∇ ⋅ 2 )ρ(r )dτ 2 4 3 2 ∫all space 4πr πR 4πε0r 4πε 0 r on the r’ coordinate? 3 Q rˆ 3 = rrˆ Since (∇ ⋅ ) = 4πδ ( ), 3 2 r 4πε R r 0 1 1 (b) outside ∇ ⋅E = 4πδ 3 (r − r′)ρ(r′)dτ ′ = ρ(r) ∫all space Φ Q 4πε 0 ε 0 ˆ ˆ E = 2 r = 2 r 4πr 4πε0r 17 18 Example 2.3 Example 2.5 Find the field due to the following: (a) an infinite sheet of A long cylinder carries a charge density that is proportional charge with surface charge density +σ; (b) two parallel to the distance from the axis: λ=ks, for some constant k. infinite sheets with charges density +σ and -σ. Find the electric field inside the cylinder? Solution: Solution: Pick up a Gaussian surface as shown in the figure. The total charge enclosed is s 2 3 Qenc = A (ks′)s′ds′dφ = πkAs ∫0 3 Q 1 E = enc = ks2 in sˆ direction ε 0 2πsA 3 19 20 How to Choose a Good Gaussian Surface? 2.2.4 The Curl of the Electric Field The electric field can be expressed in the following form Gauss’s Law is always true, but it is not always useful. 1 ˆ 1 1 Symmetry is crucial to the application of Gauss’s law. E = r ρ(r′)dτ ′ = (∇ )ρ(r′)dτ ′ 4πε ∫all space 2 4πε ∫all space There are only three kinds of symmetry that work: 0 r 0 r Curl of the electric field is Why the divergence 1. Spherical symmetry: Make your Gaussian surface a 1 1 operator doesn’t apply concentric sphere. ∇×E = (∇×(∇ ))ρ(r′)dτ ′ ∫all space on the r’ coordinate? 2. Cylindrical symmetry: Make your Gaussian surface a 4πε 0 r coaxial cylinder. Curl of gradient is always zero. ∴∇×E = 0 3. Plane symmetry: Use a Gaussian “pillbox”, which straddles the surface. The principle of superposition states that the total field is a vector sum of their individual fields E=E1+E2+… ∇×E = ∇×(E + E +") = ∇×E + ∇×E +" = 0 21 1 2 1 2 22 2.3 Electrical Potential Mechanical Analogous of Potential 2.3.1&2 Introduction to and Comments on Potential The motion of a particle with positive charge q in a uniform electric field is analogous to the motion of a particle of mass Can we apply the concept of potential, first introduced in m in uniform gravitational field near the earth. mechanics, to electrostatic system and find the law of conservation of energy? WEXT = +∆U = U f −Ui We can define an electrostatic potential energy, analogous to gravitational potential energy, and apply the law of If WEXT >0, work is done by the conservation of energy in the analysis of electrical problems. external agent on the charges. If WEXT <0, work is done on the Potential is a property of a point in space and depends only external agent by the field.