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Discuss how potential difference and electric field strength are related. Give an example.
What is the strength of the electric field in a region where the electric potential is constant?
The electric field strength is zero because electric potential differences are directly related to the field strength. If the potential difference is zero, then the field strength must also be zero.
If a proton is released from rest in an electric field, will it move in the direction of increasing or decreasing potential? Also answer this question for an electron and a neutron. Explain why.
Voltage is the common word for potential difference. Which term is more descriptive, voltage or potential difference?
Potential difference is more descriptive because it indicates that it is the difference between the electric potential of two points.
If the voltage between two points is zero, can a test charge be moved between them with zero net work being done? Can this necessarily be done without exerting a force? Explain.
What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential energy?
They are very similar, but potential difference is a feature of the system; when a charge is introduced to the system, it will have a potential energy which may be calculated by multiplying the magnitude of the charge by the potential difference.
Voltages are always measured between two points. Why?
How are units of volts and electron-volts related? How do they differ?
An electron-volt is a volt multiplied by the charge of an electron. Volts measure potential difference, electron-volts are a unit of energy.
Can a particle move in a direction of increasing electric potential, yet have its electric potential energy decrease? Explain
Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be $1.67\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-27}}\phantom{\rule{0.2em}{0ex}}\text{kg}\text{.}$
$\begin{array}{ccc}\frac{1}{2}{m}_{e}{v}_{e}^{2}\hfill & =\hfill & qV,\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{m}_{\text{H}}{v}_{\text{H}}^{2}=qV,\phantom{\rule{0.2em}{0ex}}\text{so that}\hfill \\ \frac{{m}_{e}{v}_{e}^{2}}{{m}_{\text{H}}{v}_{\text{H}}^{2}}\hfill & =\hfill & 1\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}\frac{{v}_{e}}{{v}_{\text{H}}}=42.8\hfill \end{array}$
An evacuated tube uses an accelerating voltage of 40 kV to accelerate electrons to hit a copper plate and produce X-rays. Non-relativistically, what would be the maximum speed of these electrons?
Show that units of V/m and N/C for electric field strength are indeed equivalent.
$1\phantom{\rule{0.2em}{0ex}}\text{V}=1\phantom{\rule{0.2em}{0ex}}\text{J/C;}\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}\text{J}=1\phantom{\rule{0.2em}{0ex}}\text{N}\xb7\text{m}\to 1\phantom{\rule{0.2em}{0ex}}\text{V/m}\phantom{\rule{0.2em}{0ex}}=1\phantom{\rule{0.2em}{0ex}}\text{N/C}$
What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of $1.50\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{V}$ ?
The electric field strength between two parallel conducting plates separated by 4.00 cm is $7.50\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{V}$ . (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be zero volts. What is the potential 1.00 cm from that plate and 3.00 cm from the other?
a. ${V}_{AB}=3.00\phantom{\rule{0.2em}{0ex}}\text{kV}$ ; b. ${V}_{AB}=7.50\phantom{\rule{0.2em}{0ex}}\text{kV}$
The voltage across a membrane forming a cell wall is 80.0 mV and the membrane is 9.00 nm thick. What is the electric field strength? (The value is surprisingly large, but correct.) You may assume a uniform electric field.
Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate (and 2.00 cm from the other) is 450 V? (b) What is the voltage between the plates?
a.
${V}_{AB}=Ed\to E=5.63\phantom{\rule{0.2em}{0ex}}\text{kV/m}$ ;
b.
${V}_{AB}=563\phantom{\rule{0.2em}{0ex}}\text{V}$
Find the maximum potential difference between two parallel conducting plates separated by 0.500 cm of air, given the maximum sustainable electric field strength in air to be $3.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{V/m}$ .
An electron is to be accelerated in a uniform electric field having a strength of $2.00\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{V/m}\text{.}$ (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?
a.
$\begin{array}{ccc}\text{\Delta}K\hfill & =\hfill & q\text{\Delta}V\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{V}_{AB}=Ed,\phantom{\rule{0.2em}{0ex}}\text{so that}\hfill \\ \text{\Delta}K\hfill & =\hfill & 800\phantom{\rule{0.2em}{0ex}}\text{keV;}\hfill \end{array}$
b.
$d=25.0\phantom{\rule{0.2em}{0ex}}\text{km}$
Use the definition of potential difference in terms of electric field to deduce the formula for potential difference between $r={r}_{a}$ and $r={r}_{b}$ for a point charge located at the origin. Here r is the spherical radial coordinate.
The electric field in a region is pointed away from the z-axis and the magnitude depends upon the distance s from the axis. The magnitude of the electric field is given as $E={\scriptscriptstyle \frac{\alpha}{s}}$ where $\alpha $ is a constant. Find the potential difference between points ${P}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{P}_{2}$ , explicitly stating the path over which you conduct the integration for the line integral.
One possibility is to stay at constant radius and go along the arc from ${P}_{1}$ to ${P}_{2}$ , which will have zero potential due to the path being perpendicular to the electric field. Then integrate from a to b : ${V}_{ab}=\alpha \phantom{\rule{0.2em}{0ex}}\text{ln}\left(\frac{b}{a}\right)$
Singly charged gas ions are accelerated from rest through a voltage of 13.0 V. At what temperature will the average kinetic energy of gas molecules be the same as that given these ions?
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