Fisica 3

1. (a) An Ampere is a Coulomb per second, so
84 A ⋅ h = 84

FG H

C⋅h s

IJ FG 3600 s IJ = 3.0 × 10 K H hK

5

C.

(b) The change in potential energy is ∆U = q∆V = (3.0 × 105 C)(12 V) = 3.6 × 106 J.

2. The magnitude is ∆U = e∆V = 1.2 × 109 eV = 1.2 GeV.

3. The electric field produced by an infinite sheet of charge has magnitude E = σ/2ε0, where σ is the surface charge density. The field is normal to the sheet and is uniform. Place the origin of a coordinate system at the sheet and take the x axis to be parallel to the field and positive in the direction of the field. Then the electric potential is

V = Vs −

z

x

0

E dx = Vs − Ex ,

where Vs is the potential at the sheet. The equipotential surfaces are surfaces of constant x; that is, they are planes that are parallel to the plane of charge. If two surfaces are separated by ∆x then their potentials differ in magnitude by ∆V = E∆x = (σ/2ε0)∆x. Thus, ∆x = 2ε 0 ∆ V = 2 8.85 × 10−12 C 2 N ⋅ m2 50 V . 010 × 10 C m
−6 2

c

σ

hb g = 8.8 × 10

−3

m.

4. (a) VB – VA = ∆U/q = –W/(–e) = – (3.94 × 10–19 J)/(–1.60 × 10–19 C) = 2.46 V. (b) VC – VA = VB – VA = 2.46 V. (c) VC – VB = 0 (Since C and B are on the same equipotential line).

. h c160 × 10 Ch = 2.4 × 10 N C . . (b) ∆V = E∆s = c2.4 × 10 N Chb012 mg = 2.9 × 10 V.

5. (a) E = F e = 3.9 × 10−15 N
4

c

−19

4

3

6. (a) By Eq. 24-18, the change in potential is the negative of the “area” under the curve. Thus, using the area-of-a-triangle formula, we have
V − 10 = −

z

x =2

0

E ⋅ ds =

1 2 20 2

b gb g

which yields V = 30 V. (b) For any region within 0 < x < 3 m,− E ⋅ ds is positive, but for any region for which x > 3 m it is negative. Therefore, V = Vmax occurs at x = 3 m.
V − 10 = −

z

z

x =3

0

E ⋅ ds =

1 3 20 2

b gb g

which yields Vmax = 40 V. (c) In view of our result in part (b), we see that now (to find V = 0) we are looking for some X > 3 m such that the