College Physics

# 18.4Electric Field: Concept of a Field Revisited

College Physics18.4 Electric Field: Concept of a Field Revisited

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, $F=k|q1q2|/r2F=k|q1q2|/r2 size 12{F= { ital "kq" rSub { size 8{1} } q rSub { size 8{2} } } slash {r rSup { size 8{2} } } } {}$, its magnitude is given by the equation $F=k|qQ|/r2F=k|qQ|/r2 size 12{F= { ital "kqQ"} slash {r rSup { size 8{2} } } } {}$, for a point charge (a particle having a charge $QQ size 12{Q} {}$) acting on a test charge $qq size 12{q} {}$ at a distance $rr size 12{r} {}$ (see Figure 18.20). Both the magnitude and direction of the Coulomb force field depend on $QQ size 12{Q} {}$ and the test charge $qq size 12{q} {}$.

Figure 18.20 The Coulomb force field due to a positive charge $QQ size 12{Q} {}$ is shown acting on two different charges. Both charges are the same distance from $QQ size 12{Q} {}$. (a) Since $q1q1 size 12{q rSub { size 8{1} } } {}$ is positive, the force $F1F1 size 12{F rSub { size 8{1} } } {}$ acting on it is repulsive. (b) The charge $q2q2 size 12{q rSub { size 8{2} } } {}$ is negative and greater in magnitude than $q1q1 size 12{q rSub { size 8{1} } } {}$, and so the force $F2F2 size 12{F rSub { size 8{2} } } {}$ acting on it is attractive and stronger than $F1F1 size 12{F rSub { size 8{1} } } {}$. The Coulomb force field is thus not unique at any point in space, because it depends on the test charges $q1q1 size 12{q rSub { size 8{1} } } {}$ and $q2q2 size 12{q rSub { size 8{2} } } {}$ as well as the charge $QQ size 12{Q} {}$.

To simplify things, we would prefer to have a field that depends only on $QQ size 12{Q} {}$ and not on the test charge $qq size 12{q} {}$. 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 $EE size 12{E} {}$ is defined to be the ratio of the Coulomb force to the test charge:

$E = F q , E = F q , size 12{E= { {F} over {q,} } } {}$
18.11

where $FF size 12{F} {}$ is the electrostatic force (or Coulomb force) exerted on a positive test charge $qq size 12{q} {}$. It is understood that $EE size 12{E} {}$ is in the same direction as $FF size 12{F} {}$. It is also assumed that $qq size 12{q} {}$ 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 $qq size 12{q} {}$ is simply obtained by multiplying charge times electric field, or $F = q E F = q E size 12{F=qE} {}$. Consider the electric field due to a point charge $QQ size 12{Q} {}$. According to Coulomb’s law, the force it exerts on a test charge $qq size 12{q} {}$ is $F=k|qQ|/r2F=k|qQ|/r2 size 12{F= { ital "kqQ"} slash {r rSup { size 8{2} } } } {}$. Thus the magnitude of the electric field, $EE size 12{E} {}$, for a point charge is

$E =| F q | = k | qQ qr 2 | = k |Q| r 2 . E =| F q | = k | qQ qr 2 | = k |Q| r 2 . size 12{E= { {F} over {q} } =k { { ital "qQ"} over { ital "qr" rSup { size 8{2} } } } =k { {Q} over {r rSup { size 8{2} } } } } {}$
18.12

Since the test charge cancels, we see that

$E = k |Q| r 2 . E = k |Q| r 2 . size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {}$
18.13

The electric field is thus seen to depend only on the charge $QQ size 12{Q} {}$ and the distance $rr size 12{r} {}$; it is completely independent of the test charge $qq size 12{q} {}$.

### Example 18.2

#### Calculating the Electric Field of a Point Charge

Calculate the strength and direction of the electric field $EE size 12{E} {}$ due to a point charge of 2.00 nC (nano-Coulombs) at a distance of 5.00 mm from the charge.

#### Strategy

We can find the electric field created by a point charge by using the equation $E=kQ/r2E=kQ/r2 size 12{E= { ital "kQ"} slash {r rSup { size 8{2} } } } {}$.

#### Solution

Here $Q=2.00×10−9Q=2.00×10−9 size 12{Q=2 "." "00" times "10" rSup { size 8{ - 9} } } {}$ C and $r=5.00×10−3r=5.00×10−3 size 12{r=5 "." "00" times "10" rSup { size 8{ - 3} } } {}$ m. Entering those values into the above equation gives

E = k Q r 2 = ( 8.99 × 10 9 N ⋅ m 2 /C 2 ) × ( 2.00 × 10 − 9 C ) ( 5.00 × 10 − 3 m ) 2 = 7.19 × 10 5 N/C. E = k Q r 2 = ( 8.99 × 10 9 N ⋅ m 2 /C 2 ) × ( 2.00 × 10 − 9 C ) ( 5.00 × 10 − 3 m ) 2 = 7.19 × 10 5 N/C. alignl { stack { size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {} # = $$9 "." "00" times "10" rSup { size 8{9} } N cdot m rSup { size 8{2} } "/C" rSup { size 8{2} }$$ times { { $$2 "." "00" times "10" rSup { size 8{ - 9} } C$$ } over { $$5 "." "00" times "10" rSup { size 8{ - 3} } m$$ rSup { size 8{2} } } } {} # =7 "." "20" times "10" rSup { size 8{5} } "N/C" {} } } {}
18.14

#### Discussion

This electric field strength is the same at any point 5.00 mm away from the charge $QQ size 12{Q} {}$ that creates the field. It is positive, meaning that it has a direction pointing away from the charge $QQ size 12{Q} {}$.

### Example 18.3

#### Calculating the Force Exerted on a Point Charge by an Electric Field

What force does the electric field found in the previous example exert on a point charge of $–0.250μC–0.250μC$?

#### Strategy

Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric field $E=F/qE=F/q size 12{E= {F} slash {q} } {}$ rearranged to $F=qEF=qE size 12{F= ital "qE"} {}$.

#### Solution

The magnitude of the force on a charge $q=−0.250μCq=−0.250μC size 12{q= - 0 "." "250""μC"} {}$ exerted by a field of strength $E=7.20×105E=7.20×105 size 12{E=7 "." "20" times "10" rSup { size 8{5} } } {}$ N/C is thus,

F = − qE = ( 0.250 × 10 –6 C ) ( 7.20 × 10 5 N/C ) = 0.180 N. F = − qE = ( 0.250 × 10 –6 C ) ( 7.20 × 10 5 N/C ) = 0.180 N. alignl { stack { size 12{F= ital "qE"} {} # size 12{ {}= $$"-0" "." "250" times "10" rSup { size 8{"-6"} } C$$ $$7 "." "20" times "10" rSup { size 8{5} } "N/C"$$ } {} # ="-0" "." "180"`N {} } } {}
18.15

Because $qq$ is negative, the force is directed opposite to the direction of the field.

#### Discussion

The force is attractive, as expected for unlike charges. (The field was created by a positive charge and here acts on a negative charge.) The charges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in static cling and similar situations.

### PhET Explorations

#### Electric Field of Dreams

Play ball! Add charges to the Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude.

Figure 18.21
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