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## Chapter 21 Electric Charge and Electric Field

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**+4Q**+Q 1 2 3 5 4 2R R 3R ConcepTest 21.4bElectric Force II Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q0 on the line between the two charges such that the net force on Q0 will be zero?**+4Q**+Q 1 2 3 5 4 2R R 3R ConcepTest 21.4bElectric Force II Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q0 on the line between the two charges such that the net force on Q0 will be zero? The force on Q0 due to +Q is: F = k(Q0)(Q)/R2 The force on Q0 due to +4Q is: F = k(Q0)(4Q)/(2R)2 Since +4Q is 4 times bigger than +Q, Q0 needs to be farther from +4Q. In fact, Q0 must be twice as far from +4Q, since the distance is squared in Coulomb’s law.**e**p ConcepTest 21.5cProton and Electron III 1)in the middle 2) closer to the electron’s side 3) closer to the proton’s side A proton and an electron are held apart a distance of 1 m and then let go. Where would they meet?**e**p ConcepTest 21.5cProton and Electron III 1)in the middle 2) closer to the electron’s side 3) closer to the proton’s side A proton and an electron are held apart a distance of 1 m and then let go. Where would they meet? By Newton’s 3rd law, the electron and proton feel the same force. But, since F = ma, and since the proton’s mass is much greater, the proton’s acceleration will be much smaller! Thus, they will meet closer to the proton’s original position. Follow-up: Which particle will be moving faster when they meet?**21-6 The Electric Field**For a point charge:**21-6 The Electric Field**An electric field surrounds every charge.**21-6 The Electric Field**Force on a point charge in an electric field:**21-6 The Electric Field**Example 21-6: 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.**21-6 The Electric Field**Example 21-7: 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 10-31 kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)?**-2 C**-2 C 1 2 4 3 -2 C -2 C ConcepTest 21.9bSuperposition II What is the electric field at the center of the square? • E = 0 • E = 1 • E = 2 • E = 3 • E = 4**-2 C**-2 C 1 2 4 3 -2 C -2 C ConcepTest 21.9bSuperposition II What is the electric field at the center of the square? 5) E = 0 The four E field vectors all point outward from the center of the square toward their respective charges. Because they are all equal, the net E field is zero at the center!! Follow-up: What if the upper two charges were +2 C? What if the right-hand charges were +2 C?**21-7 Electric Field Calculations for 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.**21-7 Electric Field Calculations for Continuous Charge**Distributions Example 21-9: 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).**21-7 Electric Field Calculations for Continuous Charge**Distributions Example 21-9: 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). SYMMETRY!!**21-7 Electric Field Calculations for Continuous Charge**Distributions Conceptual Example 21-10: 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.**21-7 Electric Field Calculations for Continuous Charge**Distributions Example 21-11: 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).**21-7 Electric Field Calculations for Continuous Charge**Distributions Example 21-11: 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). SYMMETRY!!**21-7 Electric Field Calculations for Continuous Charge**Distributions Example 21-12: 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.**21-7 Electric Field Calculations for Continuous Charge**Distributions Example 21-12: 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. SYMMETRY!!**21-7 Electric Field Calculations for Continuous Charge**Distributions In the previous example, if we are very close to the disk (that is, if z << R), the electric field is: This is the field due to an infinite plane of charge.**21-7 Electric Field Calculations for Continuous Charge**Distributions Example 21-13: 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).**21-8 Field Lines**The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.**21-8 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. Schematic only – fields in 3-d, not 2-d**21-8 Field Lines**Electric dipole: two equal charges, opposite in sign:**ConcepTest 21.12bElectric Field Lines II**1) 2) 3) both the same Which of the charges has the greater magnitude?**ConcepTest 21.12bElectric Field Lines II**1) 2) 3) both the same Which of the charges has the greater magnitude? The field lines are denser around the red charge, so the red one has the greater magnitude. Follow-up: What is the red/green ratio of magnitudes for the two charges?**21-8 Field Lines**The electric field between two closely spaced, oppositely charged parallel plates is constant.**21-8 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.**21-9 Electric Fields and 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.**21-9 Electric Fields and Conductors**The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move.**21-9 Electric Fields and Conductors**Conceptual Example 21-14: 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?**21-10 Motion of a Charged Particle 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.**+2Q**1 2 3 +4Q 4 d +Q 5 d ConcepTest 21.6Forces in 2D Which of the arrows best represents the direction of the net force on charge +Q due to the other two charges?**+2Q**+2Q 1 2 3 +4Q +4Q 4 d +Q 5 d ConcepTest 21.6Forces in 2D Which of the arrows best represents the direction of the net force on charge +Q due to the other two charges? The charge +2Q repels +Q toward the right. The charge +4Q repels +Q upward, but with a stronger force. Therefore, the net force is up and to the right, but mostly up. Follow-up: What would happen if the yellow charge were +3Q?**21-10 Motion of a Charged Particle in an Electric Field**Example 21-15: Electron accelerated by electric field. An electron (mass m = 9.11 x 10-31 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.**21-10 Motion of a Charged Particle in an Electric Field**Example 21-16: 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.**21-11 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.**21-11 Electric Dipoles**An electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque:**21-11 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:**21-11 Electric Dipoles**Example 21-17: Dipole in a field. The dipole moment of a water molecule is 6.1 x 10-30 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?**21-12 Electric Forces in Molecular Biology; DNA**Molecular biology is the study of the structure and functioning of the living cell at the molecular level. The DNA molecule is a double helix:**21-12 Electric Forces in Molecular Biology; DNA**The A-T and G-C nucleotide bases attract each other through electrostatic forces.**21-12 Electric Forces in Molecular Biology; DNA**Replication: DNA is in a “soup” of A, C, G, and T in the cell. During random collisions, A and T will be attracted to each other, as will G and C; other combinations will not.**Summary of Chapter 21**• Two kinds of electric charge – positive and negative. • Charge is conserved. • Charge on electron: • e = 1.602 x 10-19 C. • Conductors: electrons free to move. • Insulators: nonconductors.**Summary of Chapter 21**• 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:**Summary of Chapter 21**• 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.**Units of Chapter 22**• Electric Flux • Gauss’s Law • Applications of Gauss’s Law • Experimental Basis of Gauss’s and Coulomb’s Laws**22-1 Electric Flux**Electric flux: Electric flux through an area is proportional to the total number of field lines crossing the area.