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The picture shows a spark chamber placed on a wooden base.
A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles as they pass through the gas between the plates allows a spark to jump. The sparks are perpendicular to the plates, following electric field lines between them. The potential difference between adjacent plates is not high enough to cause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). (credit: Daderot, Wikimedia Commons)

Field and force inside an electron gun

(a) An electron gun has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. 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 μC 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 = V AB d size 12{E= { {V rSub { size 8{"AB"} } } over {d} } } {} . Once the electric field strength is known, the force on a charge is found using F = q E . Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, F = q E size 12{F`=`q`E} {} .

Solution for (a)

The expression for the magnitude of the electric field between two uniform metal plates is

E = V AB d . size 12{E= { {V rSub { size 8{"AB"} } } over {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 size 12{V rSub { size 8{"AB"} } } {} and the plate separation of 0.0400 m, we obtain

E = 25 . 0 kV 0 . 0400 m = 6 . 25 × 10 5 V/m . size 12{E= { {"25" "." "0 kV"} over {0 "." "0400 m"} } =6 "." "25"´"10" rSup { size 8{5} } " V"/m} {}

Solution for (b)

The magnitude of the force on a charge in an electric field is obtained from the equation

F = qE . size 12{F=qE} {}

Substituting known values gives

F = ( 0.500 × 10 –6 C ) ( 6.25 × 10 5 V/m ) = 0.313 N . size 12{F= \( 0 "." "500"´"10" rSup { size 8{-6} } " C" \) \( 6 "." "25"´"10" rSup { size 8{5} } " V"/m \) =0 "." "310"" N"} {}

Discussion

Note that the units are newtons, since 1 V/m = 1 N/C size 12{1" V/m"=1" N/C"} {} . The force on the charge is the same no matter where the charge is located between the plates. This is because the electric field is uniform between the plates.

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In more general situations, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because the force on a positive charge is in the direction of E and also in the direction of lower potential V . Furthermore, the magnitude of E equals the rate of decrease of V size 12{V} {} with distance. The faster V size 12{V} {} decreases over distance, the greater the electric field. In equation form, the general relationship between voltage and electric field is

E = Δ V Δ s , size 12{E= - { {ΔV} over {Δs} } } {}

where Δ s size 12{?s} {} is the distance over which the change in potential, Δ V size 12{Δ`V} {} , takes place. The minus sign tells us that E size 12{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.

Relationship between voltage and electric field

In equation form, the general relationship between voltage and electric field is

E = Δ V Δ s , size 12{E= - { {ΔV} over {Δs} } } {}

where Δ s size 12{?s} {} is the distance over which the change in potential, Δ V size 12{?`V} {} , takes place. The minus sign tells us that 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.

For continually changing potentials, Δ V size 12{Δ`V} {} and Δ s size 12{Δs} {} become infinitesimals and differential calculus must be employed to determine the electric field.

Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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