As a demonstration, from this we may calculate the potential difference between two points (
A and
B ) equidistant from a point charge
q at the origin, as shown in
[link] .
To do this, we integrate around an arc of the circle of constant radius r between
A and
B , which means we let
$d\overrightarrow{\text{l}}=r\widehat{\phi}d\phi ,$ while using
$\overrightarrow{\text{E}}=\frac{kq}{{r}^{2}}\widehat{\text{r}}$ . Thus,
However,
$\widehat{\text{r}}\xb7\widehat{\phi}=0$ and therefore
${V}_{B}-{V}_{A}=0.$
This result, that there is no difference in potential along a constant radius from a point charge, will come in handy when we map potentials.
What is the highest voltage possible between two plates?
Dry air can support a maximum electric field strength of about
$3.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{V/m}\text{.}$ Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air?
Strategy
We are given the maximum electric field
E between the plates and the distance
d between them. We can use the equation
${V}_{AB}=Ed$ to calculate the maximum voltage.
Solution
The potential difference or voltage between the plates is
(The answer is quoted to only two digits, since the maximum field strength is approximate.)
Significance
One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5-cm (1-in.) gap, or 150 kV for a 5-cm spark. This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage can cause a spark if there are spines on the surface, since sharp points have larger field strengths than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up with static electricity on dry days (
[link] ).
An electron gun (
[link] ) has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. (a) What is the electric field strength between the plates? (b) What force would this field exert on a piece of plastic with a
$0.500\text{-}\mu \text{C}$ charge that gets between the plates?
Strategy
Since the voltage and plate separation are given, the electric field strength can be calculated directly from the expression
$E=\frac{{V}_{AB}}{d}$ . Once we know the electric field strength, we can find the force on a charge by using
$\overrightarrow{\text{F}}=q\overrightarrow{\text{E}}.$ Since the electric field is in only one direction, we can write this equation in terms of the magnitudes,
$F=qE$ .
Solution
The expression for the magnitude of the electric field between two uniform metal plates is
$E=\frac{{V}_{AB}}{d}.$
Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0 kV. Entering this value for
${V}_{AB}$ and the plate separation of 0.0400 m, we obtain
Note that the units are newtons, since
$1\phantom{\rule{0.2em}{0ex}}\text{V/m}=1\phantom{\rule{0.2em}{0ex}}\text{N/C}$ . Because the electric field is uniform between the plates, the force on the charge is the same no matter where the charge is located between the plates.
In physics, mathematics, and related fields, a wave is a propagating dynamic disturbance (change from equilibrium) of one or more quantities
Abdikadir
Discuss how would orient a planar surface of area A in a uniform electric field of magnitude E0 to obtain (a) the maximum flux and (b) the minimum flux through the area.
The density of a gas of relative molecular mass 28 at a certain temperature is 0.90 K
kgmcube.The root mean square speed of the gas molecules at that temperature is 602ms.Assuming that the rate of diffusion of a gas in inversely proportional to the square root of its density,calculate the density of
A hot liquid at 80degree Celsius is added to 600g of the same liquid originally at 10 degree Celsius. when the mixture reaches 30 degree Celsius, what will be the total mass of the liquid?