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Electric Charge and Electric Field

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1 Electric Charge and Electric Field
Chapter 21 opener. This comb has acquired a static electric charge, either from passing through hair, or being rubbed by a cloth or paper towel. The electrical charge on the comb induces a polarization (separation of charge) in scraps of paper, and thus attracts them. Our introduction to electricity in this Chapter covers conductors and insulators, and Coulomb’s law which relates the force between two point charges as a function of their distance apart. We also introduce the powerful concept of electric field.

2 Static Electricity; Electric Charge and Its Conservation
Electric Charge in the Atom Insulators and Conductors Induced Charge; the Electroscope Coulomb’s Law The Electric Field Electric Field Calculations for Continuous Charge Distributions

3 Field Lines Electric Fields and Conductors Motion of a Charged Particle in an Electric Field Electric Dipoles Electric Forces in Molecular Biology: DNA Photocopy Machines and Computer Printers Use Electrostatics

4 Electric Charge and Its Conservation
Charge comes in two types, positive and negative; like charges repel and opposite charges attract. Figure Like charges repel one another; unlike charges attract. (Note color coding: positive and negative charged objects are colored rose-pink and blue-green, respectively, in this book.)

5 Electric Charge and Its Conservation
Electric charge is conserved – the arithmetic sum of the total charge cannot change in any interaction.

6 Electric Charge in the Atom
Nucleus (small, massive, positive charge) Electron cloud (large, very low density, negative charge) Figure Simple model of the atom.

7 Insulators and Conductors
Charge flows freely Metals Insulator: Almost no charge flows Most other materials Some materials are semiconductors. Figure (a) A charged metal sphere and a neutral metal sphere. (b) The two spheres connected by a conductor (a metal nail), which conducts charge from one sphere to the other. (c) The two spheres connected by an insulator (wood); almost no charge is conducted.

8 Coulomb’s Law Experiment shows that the electric force between two charges is proportional to the product of the charges and inversely proportional to the distance between them. Figure Coulomb’s law, Eq. 21–1, gives the force between two point charges, Q1 and Q2, a distance r apart.

9 Coulomb’s Law Coulomb’s law:
This equation gives the magnitude of the force between two charges.

10 Coulomb’s Law Unit of charge: coulomb, C.
The proportionality constant in Coulomb’s law is then: k = 8.99 x 109 N·m2/C2. Charges produced by rubbing are typically around a microcoulomb: 1 μC = 10-6 C.

11 Coulomb’s Law e = 1.602 x 10-19 C. Charge on the electron:
Electric charge is quantized in units of the electron charge.

12 Coulomb’s Law The proportionality constant k can also be written in terms of ε0, the permittivity of free space:

13 Coulomb’s Law The force is along the line connecting the charges, and is attractive if the charges are opposite, and repulsive if they are the same. Figure The direction of the static electric force one point charge exerts on another is always along the line joining the two charges, and depends on whether the charges have the same sign as in (a) and (b), or opposite signs (c).

14 Coulomb’s Law In its vector form, the Coulomb force is:
Figure The direction of the static electric force one point charge exerts on another is always along the line joining the two charges, and depends on whether the charges have the same sign as in (a) and (b), or opposite signs (c).

15 Coulomb’s Law Three charges in a line.
Three charged particles are arranged in a line, as shown. Calculate the net electrostatic force on particle 3 (the -4.0 μC on the right) due to the other two charges. Solution: Coulomb’s law gives the magnitude of the forces on particle 3 from particle 1 and from particle 2. The directions of the forces can be found from the geometrical arrangement of the charges (NOT by putting signs on the charges in Coulomb’s law, which is what the students will want to do). F = -1.5 N (to the left).

16 Coulomb’s Law Electric force using vector components.
Calculate the net electrostatic force on charge Q3 shown in the figure due to the charges Q1 and Q2. Figure Determining the forces for Example 21–3. (a) The directions of the individual forces are as shown because F32 is repulsive (the force on Q3 is in the direction away from Q2 because Q3 and Q2 are both positive) whereas F31 is attractive (Q3 and Q1 have opposite signs), so F31 points toward Q1. (b) Adding F32 to F31 to obtain the net force. Solution: The forces, components, and signs are as shown in the figure. Result: The magnitude of the force is 290 N, at an angle of 65° to the x axis.

17 The Electric Field The electric field is defined as the force on a small charge, divided by the magnitude of the charge: Figure Force exerted by charge Q on a small test charge, q, placed at points A, B, and C.

18 The Electric Field An electric field surrounds every charge.
Figure An electric field surrounds every charge. P is an arbitrary point.

19 The Electric Field For a point charge:

20 The Electric Field Force on a point charge in an electric field:
Figure (a) Electric field at a given point in space. (b) Force on a positive charge at that point. (c) Force on a negative charge at that point.

21 The Electric Field Electric field of a single point charge.
Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge Q = -3.0 x 10-6 C. Figure Example 21–6. Electric field at point P (a) due to a negative charge Q, and (b) due to a positive charge Q, each 30 cm from P. Solution: Substitution gives E = 3.0 x 105 N/C. The field points away from the positive charge and towards the negative one.

22 The Electric Field E at a point between two charges.
Two point charges are separated by a distance of 10.0 cm. One has a charge of -25 μC and the other +50 μC. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0 cm from the negative charge. (b) If an electron (mass = 9.11 x kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)? Figure Example 21–7. In (b), we don’t know the relative lengths of E1 and E2 until we do the calculation. Solution: a. The electric fields add in magnitude, as both are directed towards the negative charge. E = 6.3 x 108 N/C. b. The acceleration is the force (charge times field) divided by the mass, and will be opposite to the direction of the field (due to the negative charge of the electron). Substitution gives a = 1.1 x 1020 m/s2.

23 The Electric Field E above two point charges.
Calculate the total electric field (a) at point A and (b) at point B in the figure due to both charges, Q1 and Q2. Solution: The geometry is shown in the figure. For each point, the process is: calculate the magnitude of the electric field due to each charge; calculate the x and y components of each field; add the components; recombine to give the total field. a. E = 4.5 x 106 N/C, 76° above the x axis. b. E = 3.6 x 106 N/C, along the x axis.

24 The Electric Field Problem solving in electrostatics: electric forces and electric fields Draw a diagram; show all charges, with signs, and electric fields and forces with directions. Calculate forces using Coulomb’s law. Add forces vectorially to get result. Check your answer!

25 Continuous Charge Distributions
A continuous distribution of charge may be treated as a succession of infinitesimal (point) charges. The total field is then the integral of the infinitesimal fields due to each bit of charge: Remember that the electric field is a vector; you will need a separate integral for each component.

26 Continuous Charge Distributions
A ring of charge. A thin, ring-shaped object of radius a holds a total charge +Q distributed uniformly around it. Determine the electric field at a point P on its axis, a distance x from the center. Let λ be the charge per unit length (C/m). Solution: Because P is on the axis, the transverse components of E must add to zero, by symmetry. The longitudinal component of dE is dE cos θ, where cos θ = x/(x2 + a2)1/2. Write dQ = λdl, and integrate dl from 0 to 2πa. Answer: E = (1/4πε0)(Qx/[x2 + a2]3/2)

27 Continuous Charge Distributions
Charge at the center of a ring. Imagine a small positive charge placed at the center of a nonconducting ring carrying a uniformly distributed negative charge. Is the positive charge in equilibrium if it is displaced slightly from the center along the axis of the ring, and if so is it stable? What if the small charge is negative? Neglect gravity, as it is much smaller than the electrostatic forces. Solution: The positive charge is in stable equilibrium, as it is attracted uniformly by every part of the ring. The negative charge is also in equilibrium, but it is unstable; once it is displaced from its equilibrium position, it will accelerate away from the ring.

28 Continuous Charge Distributions
Long line of charge. Determine the magnitude of the electric field at any point P a distance x from a very long line (a wire, say) of uniformly distributed charge. Assume x is much smaller than the length of the wire, and let λ be the charge per unit length (C/m). Solution: By symmetry, there will be no component of the field parallel to the wire. Write dE = (λ/4πε0) (cos θ dy)/x2 + y2). This can be transformed into an integral over θ: dE = (λ/4πε0x) cos θ dθ. Integrate θ from –π/2 to + π/2. Answer: E = (1/2πε0) λ/x.

29 Continuous Charge Distributions
Uniformly charged disk. Charge is distributed uniformly over a thin circular disk of radius R. The charge per unit area (C/m2) is σ. Calculate the electric field at a point P on the axis of the disk, a distance z above its center. Solution: The disk is a set of concentric rings, and we know (from example 21-9) what the field due to a ring of charge is. Write dQ = σ 2πr dr. Integrate r from 0 to R. See text for answer.

30 Continuous Charge Distributions
In the previous example, if we are very close to the disk (that is, if z << R), the electric field is: Infinite plane This is the field due to an infinite plane of charge.

31 Continuous Charge Distributions
Two parallel plates. Determine the electric field between two large parallel plates or sheets, which are very thin and are separated by a distance d which is small compared to their height and width. One plate carries a uniform surface charge density σ and the other carries a uniform surface charge density -σ as shown (the plates extend upward and downward beyond the part shown). Solution: The field due to each plate is σ/2ε0. Between the plates the fields add, giving E = σ/ε0; outside the plates the fields cancel.

32 Field Lines The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge. Figure Electric field lines (a) near a single positive point charge, (b) near a single negative point charge.

33 Field Lines The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge. The electric field is stronger where the field lines are closer together.

34 Field Lines Electric dipole: two equal charges, opposite in sign:

35 Field Lines uniform field
The electric field between two closely spaced, oppositely charged parallel plates is constant. uniform field

36 Field Lines Summary of field lines:
Field lines indicate the direction of the field; the field is tangent to the line. The magnitude of the field is proportional to the density of the lines. Field lines start on positive charges and end on negative charges; the number is proportional to the magnitude of the charge.

37 Conductors The static electric field inside a conductor is zero – if it were not, the charges would move. The net charge on a conductor resides on its outer surface. Figure A charge inside a neutral spherical metal shell induces charge on its surfaces. The electric field exists even beyond the shell, but not within the conductor itself.

38 Conductors The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move. Figure If the electric field at the surface of a conductor had a component parallel to the surface E||, the latter would accelerate electrons into motion. In the static case, E|| must be zero, and the electric field must be perpendicular to the conductor’s surface: E = E┴.

39 Conductors Shielding, and safety in a storm.
A neutral hollow metal box is placed between two parallel charged plates as shown. What is the field like inside the box? The field inside the box is zero. This is why it can be relatively safe to be inside an automobile during an electrical storm.

40 Motion in an Electric Field
The force on an object of charge q in an electric field is given by: = q Therefore, if we know the mass and charge of a particle, we can describe its subsequent motion in an electric field.

41 Motion in an Electric Field
Electron accelerated by electric field. An electron (mass m = 9.11 x kg) is accelerated in the uniform field (E = 2.0 x 104 N/C) between two parallel charged plates. The separation of the plates is 1.5 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. (a) With what speed does it leave the hole? (b) Show that the gravitational force can be ignored. Assume the hole is so small that it does not affect the uniform field between the plates. Solution. a. The acceleration of the electron is qE/m = 3.5 x 1015 m/s2. In 1.5 cm it therefore accelerates from a speed of zero to v = 1.0 x 107 m/s. b. The electric force on the electron is qE = 3.2 x N; the gravitational force is mg = 8.9 x N. Therefore the gravitational force can safely be ignored.

42 Motion in an Electric Field
Electron moving perpendicular to . Suppose an electron traveling with speed v0 = 1.0 x 107 m/s enters a uniform electric field , which is at right angles to v0 as shown. Describe its motion by giving the equation of its path while in the electric field. Ignore gravity. Solution: The acceleration is in the vertical direction (perpendicular to the motion) and is equal to –eE/m. Then y = ½ ay2 and x = v0t; eliminating t gives the equation y = -(eE/2mv02)x2, which is a parabola.

43 Electric Dipoles An electric dipole consists of two charges Q, equal in magnitude and opposite in sign, separated by a distance . The dipole moment, p = Q , points from the negative to the positive charge. Figure A dipole consists of equal but opposite charges, +Q and –Q, separated by a distance l. The dipole moment p + Ql and points from the negative to the positive charge.

44 Electric Dipoles An electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque: Figure An electric dipole in a uniform electric field.

45 Electric Dipoles The electric field created by a dipole is the sum of the fields created by the two charges; far from the dipole, the field shows a 1/r3 dependence: Figure Electric field due to an electric dipole.

46 Electric Dipoles Dipole in a field.
The dipole moment of a water molecule is 6.1 x C·m. A water molecule is placed in a uniform electric field with magnitude 2.0 x 105 N/C. (a) What is the magnitude of the maximum torque that the field can exert on the molecule? (b) What is the potential energy when the torque is at its maximum? (c) In what position will the potential energy take on its greatest value? Why is this different than the position where the torque is maximum? Solution. a. The maximum torque occurs then the field is perpendicular to the dipole moment; τ = pE = 1.2 x N·m. b. The potential energy is zero. c. The potential energy is maximum when the dipole moment is in the opposite direction to the field; this is the largest angle through which the dipole can rotate to reach equilibrium. The torque is maximum when the moment and the field are perpendicular. This situation is analogous to a physical pendulum – its gravitational potential energy is greatest when it is poised above the axis, even though the torque is zero there.

47 Summary Two kinds of electric charge – positive and negative.
Charge is conserved. Charge on electron: e = x C. Conductors: electrons free to move. Insulators: nonconductors.

48 Summary Charge is quantized in units of e.
Objects can be charged by conduction or induction. Coulomb’s law: Electric field is force per unit charge:

49 Summary Electric field of a point charge:
Electric field can be represented by electric field lines. Static electric field inside conductor is zero; surface field is perpendicular to surface.

50 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s law. Gauss’s law involves an integral of the electric field E at each point on a closed surface. The surface is only imaginary, and we choose the shape and placement of the surface so that we can figure out the integral. In this drawing, two different surfaces are shown, both enclosing a point charge Q. Gauss’s law states that the product E·dA, where dA is an infinitesimal area of the surface, integrated over the entire surface, equals the charge enclosed by the surface Qencl divided by ε0. Both surfaces here enclose the same charge Q. Hence ∫E·dA will give the same result for both surfaces.

51 Electric Flux Gauss’s Law Applications of Gauss’s Law Experimental Basis of Gauss’s and Coulomb’s Laws

52 Electric Flux Electric flux:
Electric flux through an area is proportional to the total number of field lines crossing the area. Figure (a) A uniform electric field E passing through a flat area A. (b) E┴ = E cos θ is the component of E perpendicular to the plane of area A. (c) A┴ = A cos θ is the projection (dashed) of the area A perpendicular to the field E.

53 Electric Flux Electric flux.
Calculate the electric flux through the rectangle shown. The rectangle is 10 cm by 20 cm, the electric field is uniform at 200 N/C, and the angle θ is 30°. Solution. The flux is EA cos θ = 3.5 N·m2/C.

54 Electric Flux Flux through a closed surface:

55 Gauss’s Law The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law: This can be used to find the electric field in situations with a high degree of symmetry.

56 Gauss’s Law For a point charge, Therefore,
Solving for E gives the result we expect from Coulomb’s law: Figure A single point charge Q at the center of an imaginary sphere of radius r (our “gaussian surface”—that is, the closed surface we choose to use for applying Gauss’s law in this case).

57 Gauss’s Law Using Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives: Figure A single point charge surrounded by a spherical surface, A1, and an irregular surface,A2. Looking at the arbitrarily shaped surface A2, we see that the same flux passes through it as passes through A1. Therefore, this result should be valid for any closed surface.

58 Gauss’s Law Finally, if a gaussian surface encloses several point charges, the superposition principle shows that: Therefore, Gauss’s law is valid for any charge distribution. Note, however, that it only refers to the field due to charges within the gaussian surface – charges outside the surface will also create fields.

59 Applications of Gauss’s Law
Spherical conductor. A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere? Figure Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to determine Example 22–3. Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4πε0r2). b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must be zero. c. All the excess charge on a conductor resides on its surface, so these answers hold for a solid sphere as well.

60 Applications of Gauss’s Law
Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0). Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2). b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r03 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r03).

61 Applications of Gauss’s Law
Nonuniformly charged solid sphere. Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere. Solution: a. Consider the sphere to be made of a series of spherical shells, each of radius r and thickness dr. The volume of each is dV = 4πr2 dr. To find the total charge: Q = ∫ρE dV = 4παr05/5, giving α = 5Q/4πr05. b. The charge enclosed in a sphere of radius r will be Qr5/r05. Gauss’s law then gives E = Qr3/4πε0r05.

62 Applications of Gauss’s Law
Long uniform line of charge. A very long straight wire possesses a uniform positive charge per unit length, λ. Calculate the electric field at points near (but outside) the wire, far from the ends. Solution: If the wire is essentially infinite, it has cylindrical symmetry and we expect the field to be perpendicular to the wire everywhere. Therefore, a cylindrical gaussian surface will allow the easiest calculation of the field. The field is parallel to the ends and constant over the curved surface; integrating over the curved surface gives E = λ/2πε0R.

63 Applications of Gauss’s Law
Infinite plane of charge. Charge is distributed uniformly, with a surface charge density σ (σ = charge per unit area = dQ/dA) over a very large but very thin nonconducting flat plane surface. Determine the electric field at points near the plane. Solution: We expect E to be perpendicular to the plane, and choose a cylindrical gaussian surface with its flat sides parallel to the plane. The field is parallel to the curved side; integrating over the flat sides gives E = σ/2ε0.

64 Applications of Gauss’s Law
Electric field near any conducting surface. Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by E = σ/ε0 where σ is the surface charge density on the conductor’s surface at that point. Solution: Again we choose a cylindrical gaussian surface. Now, however, the field inside the conductor is zero, so we only have a nonzero integral over one surface of the cylinder. Integrating gives E = σ/ε0.

65 Applications of Gauss’s Law
The difference between the electric field outside a conducting plane of charge and outside a nonconducting plane of charge can be thought of in two ways: The field inside the conductor is zero, so the flux is all through one end of the cylinder. The nonconducting plane has a total charge density σ, whereas the conducting plane has a charge density σ on each side, effectively giving it twice the charge density.

66 Applications of Gauss’s Law
Conductor with charge inside a cavity. Suppose a conductor carries a net charge +Q and contains a cavity, inside of which resides a point charge +q. What can you say about the charges on the inner and outer surfaces of the conductor? Solution: The field must be zero within the conductor, so the inner surface of the cavity must have an induced charge totaling –q (so that a gaussian surface just around the cavity encloses no charge). The charge +Q resides on the outer surface of the conductor.

67 Applications of Gauss’s Law
Procedure for Gauss’s law problems: Identify the symmetry, and choose a gaussian surface that takes advantage of it (with surfaces along surfaces of constant field). Draw the surface. Use the symmetry to find the direction of E. Evaluate the flux by integrating. Calculate the enclosed charge. Solve for the field.

68 Summary Electric flux: Gauss’s law:
Gauss’s law can be used to calculate the field in situations with a high degree of symmetry. Gauss’s law applies in all situations, and therefore is more general than Coulomb’s law.

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