# 19.2 Electric potential in a uniform electric field  (Page 3/5)

 Page 3 / 5

## Section summary

• The voltage between points A and B is
$\begin{array}{c}\left(\begin{array}{c}{V}_{\text{AB}}=\mathrm{Ed}\\ E=\frac{{V}_{\text{AB}}}{d}\end{array}}\text{(uniform}\phantom{\rule{0.25em}{0ex}}E\phantom{\rule{0.25em}{0ex}}\text{- field only),}\end{array}$
where $d$ is the distance from A to B, or the distance between the plates.
• In equation form, the general relationship between voltage and electric field is
$E=\phantom{\rule{0.25em}{0ex}}–\frac{\Delta V}{\Delta s},$
where $\Delta s$ is the distance over which the change in potential, $\Delta V$ , takes place. The minus sign tells us that $\mathbf{\text{E}}$ points in the direction of decreasing potential.) The electric field is said to be the gradient (as in grade or slope) of the electric potential.

## Conceptual questions

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?

Will a negative charge, initially at rest, move toward higher or lower potential? Explain why.

## Problems&Exercises

Show that units of V/m and N/C for electric field strength are indeed equivalent.

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 $1\text{.}\text{50}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}V$ ?

The electric field strength between two parallel conducting plates separated by 4.00 cm is $7\text{.}\text{50}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{V/m}$ . (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?

(a) $3\text{.}\text{00 kV}$

(b) $\text{750 V}$

How far apart are two conducting plates that have an electric field strength of $4\text{.}\text{50}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{V/m}$ between them, if their potential difference is 15.0 kV?

(a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( $3.0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m}$ ) if the plates are separated by 2.00 mm and a potential difference of $5.0×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{V}$ is applied? (b) How close together can the plates be with this applied voltage?

(a) No. The electric field strength between the plates is $2.5×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m,}$ which is lower than the breakdown strength for air ( $3.0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m}$ ).

(b) 1.7 mm

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. Membranes are discussed in Capacitors and Dielectrics and Nerve Conduction—Electrocardiograms .) You may assume a uniform electric field.

Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. (Membranes are discussed in some detail in Nerve Conduction—Electrocardiograms .) What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m? You may assume a uniform electric field.

44.0 mV

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?

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 $3.0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m}$ .

$\text{15 kV}$

A doubly charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the plates?

An electron is to be accelerated in a uniform electric field having a strength of $2\text{.}\text{00}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m}$ . (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) $\text{800 KeV}$

(b) $\text{25.0 km}$

A few grains of table salt were put in a cup of cold water kept at constant temperature and left undisturbed. eventually all the water tasted salty. this is due to?
Aunt Faith,please i am thinking the dissolution here from the word "solution" exposed the grains of salt to be dissolved in the water.Thankyou
Junior
Junior
Aunt Faith,please i am thinking the dissolution here from the word "solution" exposed the grains of salt to be dissolved in the water.Thankyou
Junior
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20cm3 of 1mol/dm3 solution of a monobasic acid HA and 20cm3 of 1mol/dm3 solution of NaOH are mixed in a calorimeter and a temperature rise of 274K is observed. If the heat capacity of the calorimeter is 160J/K, calculate the enthalpy of neutralization of the acid.(SHCw=4.2J/g/K) Formula. (ms*cs+C)*T
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