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Transcript
Electric Charges and Forces Electric Force The electric force is one of the fundamental forces of nature. Often, as you know from experience, rubbing two objects together causes them to experience an electric force. Examples: • Unrolling plastic rap • Running a comb through hair • Rubbing rubber/plastic/glass rods with fur and silk • Walking on carpet with slippers • Consider the following experiments to help determine the nature and cause of this force: The Charge Model 4 The charge model is a theory that accounts for the observations: Charge Model: • Frictional forces, such as rubbing, add or remove something called charge to/ from an object. More vigorous rubbing produces a larger quantity of charge. • There are two kinds of electric charge: positive and negative (arbitrary names) • Objects with like charges repel • Objects with opposite charge attract • a 3rd type of charge that is attracted to both positive and negative charges has never been found. • Choice of what is a positive charge and a negative charge is arbitrary (glass rubbed with silk is defined as positive charging). • Neutral objects have an equal mixture of positive and negative charge. • The electric force is a long-range force, but decreases with increasing distance. Clicker Question Experiments with neutral object Consider a couple experiments involving a macroscopic neutral object: • How can a charged object attract a neutral, macroscopic object? Conductors, insulators, and dielectrics • Materials in which charge is free to move are conductors. • Materials in which charge isn’t free to move are insulators. Clicker Question An electroscope consists of a neutral metallic sphere connected to metal leaves via a metal rod. Why do the leaves of the electroscope repel each other when the charged rod is brought near the sphere? (metals are good conductors) 1. The leaves become negatively charged 2. The leaves become positively charged 3. The leaves are neutral, but are attracted to the charged rod Clicker question A negatively charged glass rod is brought near a spherical conductor, which is initially neutral. The spherical conductor is initially touching another spherical conductor which is also initially neutral. While the glass rod is near the conductor, the two conductors are separated. The glass rod is then removed. What are the charged states of the two conductors? 1. Left is positively charged, right is negatively charged. 2. Right is positively charged, left is negatively charged. 3. Left is positively charged, right is neutral. Coulomb’s law and the electric force • Like charges repel, and opposite charges attract, with a force that depends on • The product of the two charges • The inverse square of the distance between them • Mathematically, the electric force is described by Coulomb’s law: F�12 kq1 q2 = r̂ r2 r Here F F�12 12 is the force q1 exerts on q2 and rˆ is a unit vector pointing from q1 toward q2 . k is the Coulomb constant , 9 2 2 approximately 9.0 × 10 N ⋅ m / C . Clicker question 14 What is the angle between two strings attached to pithballs of mass m and charge Q? The superposition principle • The electric force obeys the superposition principle. • That means the force two charges exert on a third force is just the vector sum of the forces from the two charges, each treated without regard to the other charge. • The superposition principle makes it mathematically straightforward to calculate the electric forces exerted by distributions of electric charge. • The net electric force is the sum of the individual forces. Clicker Question Clicker Question Which of the arrows best represents the direction of the net force on charge +Q due to the other two charges? CT 25.11 Consider the charge configuration shown below. What is the direction of the net force on the +q charge? B +q A C h E s/2 s/2 +Q +Q Copyright Univ. of Colorado, Boulder D Clicker question A charge q is to be placed at either A or B on the mid-line of two charges +q. Will the force on q be greater at point A or at point B? A) A B) B C) Can't tell without knowing magnitude of q. 2kqQy F = 2 (a + y 2 )3/2 The Electric Field, E • The electric field at a point in space (due to a distribution of charge) is the force per unit charge that a test-charge q’ placed at that point would experience: � F � = E q� The electric field exists in space (even if no charge exists there). 22 An electron is placed at the position marked by the dot. The force on the electron is A. to the left. B. to the right. C. zero. D. There s not enough information to tell. Fields of point charges and charge distributions • Based on Coulomb’s law, the field of a point charge is radial, outward for a positive charge and inward for a negative charge. � pt E charge kq = 2 r̂ r • The superposition principle states that, for a collection of charges, the total electric field at a point is simply the summation of the individual fields due to each charge � = E � kq i � Ei = 2 r̂i ri Clicker Question The dipole: an important charge distribution • An electric dipole consists of two point charges of equal magnitude but opposite signs, held a short distance apart. • The dipole is electrically neutral, but the separation of its charges results in an electric field. • Many charge distributions, especially molecules, behave like electric dipoles. • The product of the charge and separation is the dipole moment: p = qd. • Far from the dipole, its electric field falls off as the inverse cube of the distance. 3 (E ∝ 1/r ) Field lines • Field lines show the direction of the electric field. The density of field lines is proportional the the strength of the field. • Examples: phet: charges Continuous Charge Distribution • Macroscopic objects typically contain so many excess protons or electrons (say, over 1012), it is a very good approximation to describe the distribution to be continuous. • Instead of summing over every single point charge, we integrate over the infinitesimal spatial chunks of charge. E field at P due to Δq ∆q � ∆E = k 2 r̂ r In limit that volume becomes infinitesimally small: ∆q → dq � = E � � = dE � dq � dE = k 2 r̂ r k r̂ dq 2 r � � k r̂ dq 2 r • What is dq in terms of charge density? � = E � = dE 1D 2D 3D dq=!dA dq=!dx dx dq = λdl � = E � dq = σda kλ dl r̂ E � = r2 � kσ da r̂ r2 dq = ρdV � = E � kρ dV r̂ r2 Note that, in general, r̂ changes direction during the integral! Do integral one component at a time. Example: Line Charge Let’s calculate the electric field a distance a to the right of the right end of a uniform line charge of length L and charge Q. Q � � ++++++++++ L a � = E k r̂ dq = 2 r Q λ= L Since charge distribution is uniform: Pick a coordinate system: say the x-axis along the rod with origin at left end Ex = � L 0 kλ dx kQ = ... = 2 (a + L − x) a(L + a) Note: as a becomes much bigger than L, kQ Ex → 2 a kλ dl r̂ r2 • The electric field of an infinite line of charge: • The line carries charge density λ (units are C/m): What is the vertical component, dEy, due to the little chunk of lengthT dx? 26.11e dE y=H x + + + + + + + + + + + + + + + + + +dx+ + + ++ + dEy= A) kλdx/(x2 + H 2 ) C) kλH dx/(x2 + H 2 ) E) kλx | dx/(x2 + H 2 ) 2 2 3/2 B) kλx dx/(x + H ) D) kλH dx/(x2 + H 2 )3/2 from CU Boulder Ey = � ∞ −∞ 2kλ Ey = y kλy dx (x2 + y 2 )3/2 2kλyx ∞ = � |0 y 2 (x2 + y 2 ) Figure 21.49 The electric field at point P is 1. kQ/a2 2. less than kQ/a2 but greater than zero 3. zero 4. Depends on the value of the charge at point P • The electric field on the axis of a charged ring: Ex = � k dq cos θ = 2 r cos θ = x/r = x/ � 2π 0 � k λ a dφ cos θ 2 2 x +a x2 + a2 kQx Ex = 2 (x + a2 )3/2 • The electric field on the axis of a charged ring: • Notice that since all chunks of charge contribute equally, there really isn’t a need to integrate spatially: � k dq k Qx Ex = cos θ = 2 2 r r r kQx Ex = 2 (x + a2 )3/2 • Electric field due to disk of radius R and charge Q (a distance x from center): E= E= � � dERing = R 0 � k dQRing x (x2 + r2 )3/2 k σ 2πr dr x (x2 + r2 )3/2 1 R E = 2πσkx 2 | (x + r2 )1/2 0 � x E = 2πσk 1 − √ x2 + R 2 �