Question 43

The electric field is uniform in between the plates, so the force and the acceleration are constant. Relating the potential difference to the elec- tric field gives us AV = — Ex, where x is the separation of the plates. The definition of the electric field is the force divided by the charge. Use these two pieces of information and Newton's Second Law to solve for the acceleration in terms of V. V F This shows the acceleration is directly proportional to the potential difference V.

Question 50

20.6 Force between Two Parallel Wires t 12 The magnetic field produced at the position of wire 2 due to the current in wire 1 is: 27 d The force this field exerts on 12 a length 12 of wire 2 is: 1112 172 \_ 27 d (20-7) Wire 1 Wire 2 (a) (b) Copyright 2005 prentice Hall,

12 Electric current Magnetic field at wire 2 from current in wire 1: 21tr Force on a length AL Of wire 2: F = 12ALB Force per unit length in terms Of the currents: Magnetic field F AL \_ go 1112

Question 61

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Question 62

  • Ampère-Maxwell’s Law -- A changing electric field would create a changing electric flux, which in turn would induce a magnetic field.

    {i.dS= {i.dS= μ Ienc H- μ I enc + μιοεο dt

Question 65

A solid, metal object is isolated from other charges and has charge distributed on its surface. The charge distribution is not uniform. It may be correctly concluded that the (A) electric field outside the object is zero (B) the electric field outside the object is equal to the electric field inside the object (C) the electric field outside the object is directly proportional to the distance away from the center of mass of the object (D) the electric field outside the object, but very close to the surface, is equal to the surface charge density at any location divided by the permittivity of free space (E) the electric potential on the surface of the object is not constant

For a solid, metal object the electric field inside is equal to zero. The electric field outside the object will not be zero because some excess charge is contained on the object and Gauss's Law can be applied to show that Qin would generate an electric field. This information eliminates (A) and (B). The surface of a conductor is an equipotential, so (E) is not cor- rect. The strength of the electric field should decrease as the distance away from the center increases, so (C) is not correct. (D) is correct, and we can apply Gauss's Law as shown below to the odd-shaped figure as indicated. The Qin will be equal to the surface charge density at the location times the area of the endcap of the Gaussian cylinder. This is only true very close to the surface of the object so that the Gaussian cylinder is perpendicular to the surface.

Gaussian cylinder E-field $E.dA random shape with excess positive charge

Question 68

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