RQ3-1. a) Since the object moving north in the electric field is an electron, (negative charge), by the definition of the direction of the electric field ( the direction a positive charge would go), the direction of the field must be South, since that is the direction a positive charge would go.

b) The magnitude of the field must be E = F/q = ma/q = 9.11x10-31kg ü1.6x10-16 m/s2 / 1.6x10-19 C = 9.11x10-28 N/C

(These answers come from the definition of E= kQ/r2, pg 86 & 87.)

RQ3-2. The electric field vectors would point toward the negative charge and away from the positive charge, so the negative charge would be located in the middle of all the vectors pointing toward the same spot, and the positive charge would be at the center of all the vectors whose tails point at the same spot. (See Patterns of an electric field, pg 91)

RQ3-3: (These answers come from the definition of E= kQ/r2, and the superposition principle, pg 92.)

RQ3-4: Here are a couple of ways to arrange two point; charges In order to make the electric field be zero at some location: (These answers come from the definition of E= kQ/r2, and the superposition principle, pg 92.)

In A, at the midway point between two equal positive charges, which create equal and opposite fields at that point, hence a net field at that point of zero.

In B, one would have to have one charged object positive and one negative, and , assuming two particles cannot occupy the exact same spot, the furtherest one away would have to be larger than the closer one, in order for their effects to be equal.

RQ 3-5. No object can impose a force on itself. Thus, an object cannot impose an electric field , since that would require it to apply a force to itself. (See 3.2.1 , pg 94.)

RQ3-6 (This answer comes from page 104-105), E = F/q only of q is not large enough to distrub the arrangement of charges creating the field E in the first place by polarization.

RQ3-7. (The answer to this question comes from section 3.5, pg 105-106.) The information sent by a charge that it exists and its location is transmitted at the speed of light. If it takes 45 nanoseconds to detect a change in location of the particle, it must have been 45x10-9 s (3x108m/s) = 13.5 m.

RQ3-8. (The answer for this problem comes from page 99). My answer. The reason the electric field due to an induced dipole varies inversely with d⁵ instead of varying inversely with d³ as it does for the electric field due to a permanent dipole is because the electric field for a dipole also varies directly with s, the distance between the two charges forming the dipole. This distance changes for an induced dipole, instead of remaining fixed , as it does for a permanent dipole.

Their answer: Consider moving a point charge to twice the distance from a permanent dipole, In which case the force that the dipole exerts on the point charge drops by a factor of 1/2,3 = 1/8:

Now conesider moving a point charge to twice the distance from an induced dipole, in which case the dipole separation drops by a factor of 1/4 (field of the point charge drops by 1/4), so the force on the point charge drops by a factor of (1/8)(1/4) = 1/32:

5o the key point Is that; a dipole. makes a field proportional to qs/d³, and for the induced dipole s is proportional to 1/d⁵.

RQ3-9. (The answer to this problem comes from page 101.)

RQ3-10. The relationship between force and field is that the force is caused by the field. Force is in Newtons , field has units of N/ something, where the something is causing the field.