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  • Describe a force field and calculate the strength of an electric field due to a point charge.
  • Calculate the force exerted on a test charge by an electric field.
  • Explain the relationship between electrical force (F) on a test charge and electrical field strength (E).

Contact forces, such as between a baseball and a bat, are explained on the small scale by the interaction of the charges in atoms and molecules in close proximity. They interact through forces that include the Coulomb force    . Action at a distance is a force between objects that are not close enough for their atoms to “touch.” That is, they are separated by more than a few atomic diameters.

For example, a charged rubber comb attracts neutral bits of paper from a distance via the Coulomb force. It is very useful to think of an object being surrounded in space by a force field . The force field carries the force to another object (called a test object) some distance away.

Concept of a field

A field is a way of conceptualizing and mapping the force that surrounds any object and acts on another object at a distance without apparent physical connection. For example, the gravitational field surrounding the earth (and all other masses) represents the gravitational force that would be experienced if another mass were placed at a given point within the field.

In the same way, the Coulomb force field surrounding any charge extends throughout space. Using Coulomb’s law, F = k | q 1 q 2 | / r 2 size 12{F= { ital "kq" rSub { size 8{1} } q rSub { size 8{2} } } slash {r rSup { size 8{2} } } } {} , its magnitude is given by the equation F = k | qQ | / r 2 size 12{F= { ital "kqQ"} slash {r rSup { size 8{2} } } } {} , for a point charge    (a particle having a charge Q size 12{Q} {} ) acting on a test charge     q size 12{q} {} at a distance r size 12{r} {} (see [link] ). Both the magnitude and direction of the Coulomb force field depend on Q size 12{Q} {} and the test charge q size 12{q} {} .

In part a, two charges Q and q one are placed at a distance r. The force vector F one on charge q one is shown by an arrow pointing toward right away from Q. In part b, two charges Q and q two are placed at a distance r. The force vector F two on charge q two is shown by an arrow pointing toward left toward Q.
The Coulomb force field due to a positive charge Q size 12{Q} {} is shown acting on two different charges. Both charges are the same distance from Q size 12{Q} {} . (a) Since q 1 size 12{q rSub { size 8{1} } } {} is positive, the force F 1 size 12{F rSub { size 8{1} } } {} acting on it is repulsive. (b) The charge q 2 size 12{q rSub { size 8{2} } } {} is negative and greater in magnitude than q 1 size 12{q rSub { size 8{1} } } {} , and so the force F 2 size 12{F rSub { size 8{2} } } {} acting on it is attractive and stronger than F 1 size 12{F rSub { size 8{1} } } {} . The Coulomb force field is thus not unique at any point in space, because it depends on the test charges q 1 size 12{q rSub { size 8{1} } } {} and q 2 size 12{q rSub { size 8{2} } } {} as well as the charge Q size 12{Q} {} .

To simplify things, we would prefer to have a field that depends only on Q size 12{Q} {} and not on the test charge q size 12{q} {} . The electric field is defined in such a manner that it represents only the charge creating it and is unique at every point in space. Specifically, the electric field E size 12{E} {} is defined to be the ratio of the Coulomb force to the test charge:

E = F q , size 12{E= { {F} over {q,} } } {}

where F size 12{F} {} is the electrostatic force (or Coulomb force) exerted on a positive test charge q size 12{q} {} . It is understood that E size 12{E} {} is in the same direction as F size 12{F} {} . It is also assumed that q size 12{q} {} is so small that it does not alter the charge distribution creating the electric field. The units of electric field are newtons per coulomb (N/C). If the electric field is known, then the electrostatic force on any charge q size 12{q} {} is simply obtained by multiplying charge times electric field, or F = q E size 12{F=qE} {} . Consider the electric field due to a point charge Q size 12{Q} {} . According to Coulomb’s law, the force it exerts on a test charge q size 12{q} {} is F = k | qQ | / r 2 size 12{F= { ital "kqQ"} slash {r rSup { size 8{2} } } } {} . Thus the magnitude of the electric field, E size 12{E} {} , for a point charge is

Practice Key Terms 3

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Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
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