Q#23.40
Certain sharks can detect an electric field as 1.0 $\mu$V/m weak as To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5-V AA battery across these plates, how far apart would the plates have to be?
Answer:
$V_{ab}$ = Ed for parallel plates.
$d=\frac{V_{ab}}{E}=\frac{1.5 \ V}{1.0 \times 10^{-6} \ V/m}$
$d=1.5 \times 10^6$ m
The plates would have to be nearly a thousand miles apart with only a AA battery across them! This is a small field!
Q#23.41
(a) Show that for a spherical shell of radius R, that has charge q distributed uniformly over its surface, is the same as V for a solid conductor with radius R and charge q.
(b) You rub an inflated balloon on the carpet and it acquires a potential that is 1560 V lower than its potential before it became charged. If the charge is uniformly distributed over the surface of the balloon and if the radius of the balloon is 15 cm, what is the net charge on the balloon?
(c) In light of its 1200-V potential difference relative to you, do you think this balloon is dangerous? Explain.
Answer:
Consider the electric field outside and inside the shell and use that to deduce the potential.
(a) The electric field outside the shell is the same as for a point charge at the center of the shell, so the potential outside the shell is the same as for a point charge:
$V=\frac{q}{4\pi \epsilon_0 r}$ for r > R
The electric field is zero inside the shell, so no work is done on a test charge as it moves inside the shell and all points inside the shell are at the same potential as the surface of the shell:
$V=\frac{q}{4\pi \epsilon_0 R}$ for r $\leqslant$ R
(b) $V=\frac{kq}{R}$ so $q=\frac{RV}{k}$
$q=\frac{RV}{k}=\frac{(0.15 \ m)(-1200 \ V)}{8.988 \times 10^9 \ N.m^2/C^2}=-20$ nC
(c) No, the amount of charge on the sphere is very small. Since U = qV the total amount of electric energy stored on the balloon is only
$U=(-20 \times 10^{-9} \ C)(1200 V)=2.4 \times 10^{-6}$ J
Q#23.42
(a) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 1.50 kV?
(b) What is the potential of the sphere’s surface relative to infinity?
Answer:
The electric field is zero inside the sphere, so the potential is constant there. Thus the potential at the center must be the same as at the surface, where it is equivalent to that of a point-charge.
At the surface, and hence also at the center of the sphere, the potential is that of a point-charge,
$V=\frac{Q}{4\pi \epsilon_0 R}$
(a) Solving for Q and substituting the numbers gives
$Q=4\pi \epsilon_0RV$
$Q=4\pi (8.85 \times 10^{-12} \ C^2/N.m^2)(0.125 \ m)(1500 V)$
$Q=2.08 \times 10^{-8} \ C$ = 20.8 nC
(b) Since the potential is constant inside the sphere, its value at the surface must be the same as at the center, 1.50 kV.
The electric field inside the sphere is zero, so the potential is constant but is not zero.
Q#23.43
The electric field at the surface of a charged, solid, copper sphere with radius 0.200 m is 3800 N/C directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?
Answer:
The potential of a solid conducting sphere is the same at every point inside the sphere and is equal to its value $V=\frac{q}{4\pi \epsilon_0 R}$ at the surface.
Use the given value of E to find q.
For negative charge the electric field is directed toward the charge.
For points outside this spherical charge distribution the field is the same as if all the charge were concentrated at the center.
$E=\frac{|q|}{4\pi \epsilon_0r^2}$
$|q|=4\pi \epsilon_0Er^2$
$|q|=4\pi(8.85 \times 10^{-12} \ C^2/N.m^2)(3800 N/C)(0.200 \ m)^2$
$|q|=1.69 \times 10^{-8}$ C = 16.9 nC
Since the field is directed inward, the charge must be negative. The potential of a point charge, taking ∞ as zero, is
$V=\frac{q}{4\pi \epsilon_0 r}$
$V=\frac{(8.988 \times 10^9 N.m^2/C^2)(-1.69 \times 10^{-8} \ C)}{0.200 \ m}=-760 \ V$ at the surface of the sphere.
Since the charge all resides on the surface of a conductor, the field inside the sphere due to this symmetrical distribution is zero. No work is therefore done in moving a test charge from just inside the surface to the center, and the potential at the center must also be –760 V.
Inside the sphere the electric field is zero and the potential is constant.
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