ANSWER TO CHAPTER 23 HOMEWORK

ANSWER TO CHAPTER 23 HOMEWORK

1. We find the potential difference from the work-energy principle:

W_ab = DK + DU = DK + q(V_b - V_a)

8.00 x 10^-4J = 2.10 x 10^-4J + (-8.10 x 10^-6C)(V_b - V_a),

which gives V_b - V_a = -72.8V, or V_a- V_b = +72.8V.

2. The potential difference between two points in an electric field is found from

DV = -ò E × dl.

(a) For V_BA we have

V_BA = - ò_A^B E × dl = - ò_A^B (-300N/C)i × dyj = 0.

^{(b) For V_CB we have


V_CB = - ò_B^CE ×
dl = - ò_B^C (-300N/C)i × dxi = ò_4m^-3m
(300N/C)dx


= (300N/C)(-3m -4m) = -2100V.

          (c) For V_CAwe have


V_CA
= - ò_C^AE ×
dl = - ò_A^C(-300N/C)i ×
(dxi + dyj) = ò_4m^-3m
(300N/C)dx


= (300N/C)(-3m -4m) = -2100V.


           Note
that V_CA= V_CB+ V_BA.

3. The radial electric field of the long wire is


E = l/2pe_or.

        We find the
potential difference from


V_b- V_a = - ò_Ra^Rb
E × dl = - ò_Ra^Rb
Edr = - ò_Ra^Rb(ldr)/(2pe_or) =
- (l/2pe_o)ln(R_b/R_a) =
(l/2pe_o)ln(R_a/R_b).

4. We find the electric potentials of the
stationary charges at the initial and final points:


V_a = (1/4pe_o)[Q₁/r_1a)
+ (Q₂/r_2a)


= (9.0 x 10⁹N.m²/C²){[(25 x 10^-6C)/(0.030m)]
+ [(25 x 10^-6C)/(0.030m)]} = 1.50 x 10⁷V.


V_b = (1/4pe₀)[(Q₁/r_1b)+ (Q₂/r_2b)]


= (9.0 x 10⁹N.m²/C²){[(25 x 10^-6C)/(0.040m]
+ [(25 x 10^-6C)/(0.020m)]} = 1.69 x 10⁷V.

         Because there
is no change in kinetic energy, we have


W_a-b = DK + DU
= 0 + q(V_b - V_a)


= (0.10 x10^-6C)(1.69 x 10⁷V - 1.50 x 10⁷V)
= +0.19J.

5 We choose a ring of radius r and
width dr for a differential element, with charge dq = s2prdr.
The potential of this element on the axis a distance x from the
ring is

dV = dq/4pe_o(x² + r²)^1/2

= s2prdr/4pe_o(x²+ r²)^1/2= srdr/2e_o(x²
+ ²)^1/2
        We integrate to get the
potential:

V = ò_R1^R2
(srdr)/(2e_o(x²
+ r²)^1/2 = s/2e_o(x²
+ r²)^1/2|_R1^R2

= s/2eo[(x²
+ R₂²)^1/2 - (x²
+ R₁²)^1/2].

6.

7.  (a) With the distance
measured from the center of the dipole, we find the potential from each charge


V_O = Q_O/4pe_or_O


= (9.0 x 10⁹N.m²/C²)(-6.6 x 10^-20C)/(9.0
x 10^-10m - 0.6 x 10^-10m) = -0.707V.


V_C = Q_C/4pe_or_C


= (9.0 x 10⁹N.m²/C²)(+6.6 x 10^-20C)/(9.0
x 10^-10m + 0.6 x 10 ^-10m) = +0.619V.


Thus the total potential is


V = V_O + V_C = -0.707V + 0.619V = -0.088V.

          (b) The
percent error introduced by the dipole approximation is

% error = (100)(0.089V - 0.088V)/(0.088V) = 1%.

8. (a) The kinetic energy of the electron (q
= -e) is


K_e = -qV_BA = -(-e)V_BA = eV_BA.


The kinetic energy of the proton (q = +e) is


K_P= -qV_AB = -(+e)(-V_BA) =
eV_BA= 2.0keV

          (b) We
find the ratio of their speeds, starting from rest, from

1/2m_ev_e² = 1/2m_pv_p²,
or v_e/v_p = (m_p/m_e)^1/2
= [(1.67 x 10^-27kg)/(9.11 x 10^-31kg)]^1/2 = 42.8.

9*. The field outside the cylinder is the same
as that of a long wire. we find the equivalent linear charge density from the
charge on the length L:


Q = s2pR₀L
= lL, which gives l
= s2pR₀.

(a) The radial electric field outside the cylinder is


E = l/2pe₀r
= s2pR₀/2pe₀r
= sR₀/e₀r.

We find the potential difference from


V - V₀= - ò_R0^r
E · dl = - ò_R0^r
sR₀dr/e₀r


= -(sR₀/e₀)ln(r/V₀), or


V = V₀ + (sR₀/e₀)ln(R₀/r),
r > R₀.

(b) The electric field inside the cylinder is zero, so
the potential inside is constant and equal to the potential at the surface: V
= V₀, r < V₀.

(c) From the result in part (a) we see that the
potential at r = ¥ is undefined. V ¹
0, because there would be
charge at infinity for an infinite cylinder.

10*. We choose a differential element of the rod at
position x', length dx', and charge dq = ldx'
= ax'dx'.
(a) From the diagram, we see that r² = x'²
+ y². The potential on the y-axis from the
differential element is

dV = (1/4pe₀)dq/r
= ax'dx'/4pe₀r.
The potential from the rod is

V = a/4pe₀ò_x'=-L^x'=Lx'dx'/r
= a/4pe₀ò_x'=-L^x'=Lx'dx'/Öx'²
+ y²

= a/4pe₀Öx'²
+ y²|_-L^L = a/4pe₀[Ö(L²
+ y²) - Ö(L²
+ y²)

= 0.
This is expected because the potential from the negative charge on the
left half pf tje rp. is balanced by the potential from the positive charge
on the right half of the rod.
(b) From the diagram, we see that r = x - x'.
The potential on the x-axis from the differential element is

V = (1/4pe₀)dq/r =
ax'dx'/4pe₀r.
The potential from the rod is

V = a/4pe₀ò_x'=-L^x'=Lx'dx'/r
= a/4pe₀ò_-L^Lx'dx'/(x
- x')

= a/4pe₀[x - x'
- xln(x - x')]|_-L^L

= a/4pe₀{-L - L - xln[(x
- L)/(x - L)]}

= a/4pe₀{xln[(x
- L)/(x - L)] - 2L}, x > L.
Note that we have used a length of 2L for the rod.

11*. Because the field is uniform, the magnitudes of the forces
on the charges of the dipole will be equal:

F₊ = F_- = QE.
If the separation of the charges is l, the dipole moment will be
p =Ql. If we choose the center of the dipole for the axis of
rotation, both forces create a CCW torque with a net torque of

t = F₊(1/2l)sinf
+ F_-(1/2l)sinf
= 2QE(1/2l)sinf = pEsinf.
Because the forces are in opposite directions, the not force is zero.
If the field is nonuniform, there would be a torque produced by
the average field. The magnitudes of the forces would not be the same,so
there would be a resultant force
that would cause a translation of the dipole.

12*. We assume that r² >> l².
so we can use the potential produces by a dipole. We choose the coordinate
system shown, so that x = rcosq
and y = sinq. Thus the potential
produced by p is

V = pcosq/4pe₀r
= (p/4pe₀)x/(x²
+ y²)^3/2.
We find the components of E from

E_x = - ¶V/¶x
= -(p/4pe₀)¶[x/(x²
+ y²)^3/2]/¶x

= -(p/4pe₀)[1/(x²
+ y²)^3/2 - 3x²/(x²
+ y²)^5/2]

= (p/4pe₀)[(2x²
- y²)/(x²
+ y²)^5/2

= (p/4pe₀)(2cos²q
- sin²q)/r³.

E_y = - ¶V/¶y
= -(p/4pe₀)¶[x/(x²
+ y²)^3/2]/¶y

= -(p/4pe₀)[-3xy/(x²
+ y²)^5/2]

= (p/4pe₀)[3xy/(x²
+ y²)^5/2]

= (3psinqcosq)/4pe₀r³.

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