Q#23.30
An infinitely long line of charge has linear charge density 5.00 $\times 10^{-12}$ C/m. A proton (mass 1.67 $\times 10^{-27}$ kg, charge +1.60 $\times 10^{-19} \ C$) is 18.0 cm from the line and moving directly toward the line at 1.50 $\times 10^3$ m/s.
(a) Calculate the proton’s initial kinetic energy.
(b) How close does the proton get to the line of charge?
Answer:
For a line of charge $V_a-V_b=\frac{\lambda}{2\pi \epsilon_0}ln (r_b/r_a)$
Apply conservation of energy to the motion of the proton.
Let point a be 18.0 cm from the line and let point b be at the distance of closest approach, where $K_b$ = 0.
(a) $K_a=\frac{1}{2}mv^2=\frac{1}{2}(1.67 \times 10^{-27} \ kg)(1.50 \times 10^3 \ m/s)^2$
$K_a=1.88 \times 10^{-21}$ J
(b) $K_a+U_a=K_b+U_b$ with U = qV.
$K_a+qV_a=K_b+qV_b$
$V_a-V_b=\frac{K_b-K_a}{q}$
$V_a-V_b=\frac{-1.88 \times 10^{-21} \ J}{1.60 \times 10^{-19} \ C}=-0.01175$ V
$-0.01175 \ V=\frac{\lambda}{2\pi \epsilon_0}ln (r_b/r_a)$
$r_b=r_a \ exp \left(\frac{2\pi \epsilon_0(-0.01175 \ V)}{\lambda}\right)$
$r_b=(0.180 \ m) \ exp \left(\frac{2\pi \epsilon_0(-0.01175 \ V)}{5.00 \times 10^{-12} \ C/m}\right)=0.158 \ m$
The potential increases with decreasing distance from the line of charge. As the positively charged proton approaches the line of charge it gains electrical potential energy and loses kinetic energy.
Q#23.31
A very long wire carries a uniform linear charge density $\lambda$. Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed 2.50 cm from the wire and the other probe is 1.00 cm farther from the wire, the meter reads 575 V.
(a) What is $\lambda$?
(b) If you now place one probe at 3.50 cm from the wire and the other probe 1.00 cm farther away, will the voltmeter read 575 V? If not, will it read more or less than 575 V? Why?
(c) If you place both probes from the wire but 17.0 cm from each other, what will the voltmeter read?
Answer:
The voltmeter measures the potential difference between the two points. We must relate this quantity to the linear charge density on the wire.
For a very long (infinite) wire, the potential difference between two points is
$\Delta V=\frac{\lambda}{2\pi \epsilon_0}ln (r_b/r_a)$
(a) Solving for λ gives
(b) The meter will read less than 575 V because the electric field is weaker over this 1.00-cm distance than it was over the 1.00-cm distance in part (a).
(c) The potential difference is zero because both probes are at the same distance from the wire, and hence at the same potential.
Since a voltmeter measures potential difference, we are actually given ΔV, even though that is not stated explicitly in the problem.
Q#23.32
A very long insulating cylinder of charge of radius 2.50 cm carries a uniform linear density of 15.0 nC/m. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 175 V?.
Answer:
The voltmeter reads the potential difference between the two points where the probes are placed. Therefore we must relate the potential difference to the distances of these points from the center of the cylinder. For points outside the cylinder, its electric field behaves like that of a line of charge.
Using $\Delta V=\frac{\lambda}{2\pi \epsilon_0}ln (r_b/r_a)$ and solving for $r_b$, we have
$r_b=r_ae^{2\pi \epsilon \Delta V/ \lambda}$
The exponent is
$\frac{\left(\frac{1}{2 \times 9 \times 10^9 \ N.m^2/C^2}\right)(175 \ V)}{15.0 \times 10^{-9} \ C/m}=0.648$
which gives
$r_b=(2.50 \ cm)e^{0.648}$ = 4.78 cm
Since a voltmeter measures potential difference, we are actually given ΔV, even though that is not stated explicitly in the problem. We must also be careful when using the formula for the potential difference because each r is the distance from the center of the cylinder, not from the surface.
Q#23.33
A very long insulating cylindrical shell of radius carries charge of linear density 8.50 $\mu$C/m spread uniformly over its outer surface. What would a voltmeter read if it were connected between (a) the surface of the cylinder and a point 4.00 cm above the surface, and (b) the surface and a point 1.00 cm from the central axis of the cylinder?
Answer:
Since a voltmeter measures potential difference, we are actually given ΔV, even though that is not stated explicitly in the problem. We must also be careful when using the formula for the potential difference because each r is the distance from the center of the cylinder, not from the surface.
The potential difference is $\Delta V=\frac{\lambda}{2\pi \epsilon_0}ln (r_b/r_a)$
(a) Substituting numbers gives
$\Delta V=\frac{8.50 \times 10^{-6} \ C/m}{2 \times 9.00 \times 10^9 \ N.m^2/C^2}ln (10.0 \ cm/6.00 \ cm)=78.2 \times 10^3$ V
(b) E = 0 inside the cylinder, so the potential is constant there, meaning that the voltmeter reads zero.
Caution! The fact that the voltmeter reads zero in part (b) does not mean that V = 0 inside the cylinder. The electric field is zero, but the potential is constant and equal to the potential at the surface.
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