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Chapter 22: Electric Fields. Introduction: What do we really want to get out of chapter 23? Review: vectors!. 22-1 Charges and Forces: A closer look: Why fields? To explain “ action at a distance ” !!. 22-2 The Electric Field ( E ):
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Chapter 22: Electric Fields • Introduction: • What do we really want to get out of chapter 23? • Review: vectors! 22-1 Charges and Forces: A closer look: Why fields? To explain “action at a distance”!!
22-2 The Electric Field (E): To find the field (E) at a location (P) due to a charge q, calculate: E = F/q0 where q0 is a “test charge” places at P. The units of E is N/C. Typical fields: At hydrogen nucleus surface: 1011 N/C At surface of photocopier drum: 105 N/C Near a charged comb: 103 N/C In copper wire circuits: 10-2 N/C
22-3 Electric Field Lines: Are merely to visualize patterns in electric fields. 1- The direction of the tangent to the field lines at any point is the direction of E at that point. 2- The number of lines per unit area is proportional to the magnitude of E.
Example: 1- A single (positive or negative) point charge. 2- A charged conducting sphere. 3- Two (like or unlike) point charges. 4- A charged slab (or sheet) [finite or infinite] 5- A charged line (or cylinder) [short or long] Note that electric field lines originate at positive charges (or at infinity) and terminate at negative charges (or at infinity).
No two fields lines can cross or touch. Field lines are not material objects. Although the lines are drawn as discrete, the field itself is continuous.
22-4 Electric Field Due to a Point Charge: For a charge q: |F| = k |q| |q0|/r2 Therefore, E = k |q|/r2 point away from q if q is positive E = k |q|/r2 point toward q if q is negative If there are many charges q1, q2, q3, …, qn: Use “superposition”: the net field is the vector sum of the fields due to each of the n charges.
E = F/q0 =E1 + E2 + … + En Away from qj if qj is positive; toward qj if qj is negative! Check point1:
22-5 E Due to an Electric Dipole: If we want to find the electric field due to a dipole at a point which is a distance z away from the midpoint between the two dipole charges, then: Edipole = E+ + E- p is called the electric dipole moment; its magnitude is: q · d, where d is the distance between the two charges. p is a vector that points from the negative charge to the positive charge.
Along the dipole axis, when z >> d, one finds that: Edipole = 2k p/z3 and is directed along the dipole axis. Interaction: prove it!! Note that, for large distances, Edipole ~ p (not q alone or d alone); also E ~ 1/r3 not 1/r2!! What are the SI units of p?
22-8 A point charge in E: When a point charge q is places in a field E, the force experienced by this charge is: F = q E Applications: (see H&R pg. 533) 1- Millikan oil-drop experiment: q = n e 2- Ink-jet printing 3- Volcanic lightning
22-9 A dipole in in E: H2O ~ an electric dipole… why? • For a uniform E: • Fnet = 0 • But the torque (t): • = p x E The torque tends to rotate p into the direction of E!!
Taking qr = 90o, U(qr) = 0 Therefore: U(q) = - p E cos q = - p · E What is the potential energy in a dipole field? The change in potential energy is the negative of the work done by field on the dipole. Compare with gravitational potential energy
Therefore, the work done by the field on the dipole is: W = -DU = Ui– Uf However, the applied work by an external agent is: Wagent = Uf– Ui Solve problem:
