# Capitulo 4 fisica iii

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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 chargedensity. 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 planesthat 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 Cand 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

bgb 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“area” from x = 3 m to x = X is 40 V. Using the formula for a triangle (3 < x < 4) and a rectangle (4 < x < X), we require
1 1 20 + X − 4 20 = 40 . 2

b gb g b

gb g

Therefore, X = 5.5 m.

7. (a) The work done by the electric field is (in SI units) W=
f i

qσ q0 E ⋅ ds = 0 2ε 0

d 0

q0σ d (1.60 ×10−19 )(5.80 ×10−12 )(0.0356) = = 1.87 ×10−21 J. dz = −12 2ε 0 2(8.85 ×10 )

(b) SinceV – V0 = –W/q0 = –σz/2ε0, with V0 set to be zero on the sheet, the electric potential at P is (in SI units)

σz (5.80 ×10−12 )(0.0356) V =− =− = −1.17 ×10−2 V. −12 2ε 0 2(8.85 ×10 )

8. We connect A to the origin with a line along the y axis, along which there is no change of potential (Eq. 24-18: E ⋅ ds = 0 ). Then, we connect the origin to B with a line along

z

the x axis, along whichthe change in potential is
∆V = −

z

x =4

0

E ⋅ ds = −4.00

z

4

0

x dx = −4.00

FG 4 IJ H 2K
2

which yields VB – VA = –32.0 V.

9. (a) The potential as a function of r is (in SI units) V ( r ) = V (0) − =−
r 0

E ( r )dr = 0 −

r 0

qr qr 2 dr = − 4πε 0 R 3 8πε 0 R 3

(8.99 ×109 )(3.50 ×10−15 )(0.0145) 2 = −2.68 ×10−4 V. 2(0.0231)3

(b) Since ∆V = V(0) –V(R) = q/8πε0R, we have (in SI units) (8.99 ×109 )(3.50 ×10−15 ) V ( R) = − =− = −6.81×10−4 V. 8πε 0 R 2(0.0231) q

10. The charge is

q = 4πε 0 RV =

(10m) (−1.0V) 8.99 ×10 N ⋅ m
9

2

/C

2

= −1.1×10−9 C.

11. (a) The charge on the sphere is q = 4 πε 0 VR = (200 V)(0.15 m) = 3.3 × 10 −9 C. N ⋅ m2 8.99 × 10 9 C2

(b) The (uniform) surface charge density (charge divided bythe area of the sphere) is

σ=

q 3.3 × 10 −9 C = = 12 × 10 −8 C / m 2 . . 4 πR 2 4 π 015 m 2 .

b

g

12. (a) The potential difference is VA − VB = q 4πε 0 rA − q 4πε 0 rB = (1.0 ×10
−6

C)

N ⋅ m2 8.99 ×10 C2
9

1 1 − 2.0 m 1.0 m

= −4.5 ×103 V. (b) Since V(r) depends only on the magnitude of r , the result is unchanged.

13. First, we observe that V (x) cannot be...