185. Electric Field: Concept of a Field RevisitedLearning Objectives
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 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 Concept of a FieldA 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, $F=k\mathrm{}{q}_{1}{q}_{2}\mathrm{}/{r}^{2}$, its magnitude is given by the equation
$F=k\mathrm{}\mathrm{qQ}\mathrm{}/{r}^{2}$, for a To simplify things, we would prefer to have a field that depends only on $Q$ and not on the test charge $q$. 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 $E$ is defined to be the ratio of the Coulomb force to the test charge: $$\mathbf{\text{E}}=\frac{\mathbf{\text{F}}}{q},$$where $\mathbf{\text{F}}$ is the electrostatic force (or Coulomb force) exerted on a positive test charge $q$. It is understood that $\mathbf{\text{E}}$ is in the same direction as $\mathbf{\text{F}}$. It is also assumed that $q$ 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 $q$ is simply obtained by multiplying charge times electric field, or $\mathbf{\text{F}}=q\mathbf{\text{E}}$. Consider the electric field due to a point charge $Q$. According to Coulomb’s law, the force it exerts on a test charge $q$ is $F=k\mathrm{}\mathrm{qQ}\mathrm{}/{r}^{2}$. Thus the magnitude of the electric field, $E$, for a point charge is $$E=\left\frac{F}{q}\right=k\frac{\text{qQ}}{{\mathrm{qr}}^{2}}=k\frac{\leftQ\right}{{r}^{2}}.$$Since the test charge cancels, we see that $$E=k\frac{\leftQ\right}{{r}^{2}}.$$The electric field is thus seen to depend only on the charge $Q$ and the distance $r$; it is completely independent of the test charge $q$. Example 1: Calculating the Electric Field of a Point ChargeCalculate the strength and direction of the electric field $E$ due to a point charge of 2.00 nC (nanoCoulombs) 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 $E=\text{kQ}/{r}^{2}$. Solution Here $Q=2\text{.}\text{00}\times {\text{10}}^{9}$ C and $r=5\text{.}\text{00}\times {\text{10}}^{3}$ m. Entering those values into the above equation gives $$\begin{array}{lll}E& =& k\frac{Q}{{r}^{2}}\\ & =& (\text{8.99}\times {\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{m}}^{2}{\text{/C}}^{2})\times \frac{(\text{2.00}\times {\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{C})}{(\text{5.00}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m}{)}^{2}}\\ & =& \text{7.19}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C.}\end{array}$$Discussion This Example 2: Calculating the Force Exerted on a Point Charge by an Electric FieldWhat force does the electric field found in the previous example exert on a point charge of $\mathrm{\u20130.250}\phantom{\rule{0.25em}{0ex}}\mu \text{C}$? 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 $\mathbf{\text{E}}=\mathbf{\text{F}}/q$ rearranged to $\mathbf{\text{F}}=q\mathbf{\text{E}}$. Solution The magnitude of the force on a charge $q=0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\text{\mu C}$ exerted by a field of strength $E=7\text{.}\text{20}\times {\text{10}}^{5}$ N/C is thus, $$\begin{array}{lll}F& =& \text{qE}\\ & =& (\text{0.250}\times {\text{10}}^{\text{\u20136}}\phantom{\rule{0.25em}{0ex}}\text{C})(7.20\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C})\\ & =& \text{0.180 N.}\end{array}$$Because $q$ 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 DreamsPlay 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. Section Summary
Conceptual QuestionsExercise 1Why must the test charge $q$ in the definition of the electric field be vanishingly small? Exercise 2Are the direction and magnitude of the Coulomb force unique at a given point in space? What about the electric field? Problem ExercisesExercise 1What is the magnitude and direction of an electric field that exerts a $2\text{.}\text{00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}$ upward force on a $\mathrm{\u20131.75}\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ charge? Exercise 2What is the magnitude and direction of the force exerted on a $3.50\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ charge by a 250 N/C electric field that points due east? Show/Hide Solution Solution$8\text{.}\text{75}\times {\text{10}}^{4}$ N Exercise 3Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff). Exercise 4(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0 m? Show/Hide Solution Solution(a) $6\text{.}\text{94}\times {\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{C}$ (b) $6\text{.}\text{25}\phantom{\rule{0.25em}{0ex}}\text{N/C}$ Exercise 5Calculate the initial (from rest) acceleration of a proton in a $5\text{.}\text{00}\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{N/C}$ electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the ProblemSolving Strategy for electrostatics. Exercise 6(a) Find the direction and magnitude of an electric field that exerts a $4\text{.}\text{80}\times {\text{10}}^{\text{17}}\phantom{\rule{0.25em}{0ex}}\text{N}$ westward force on an electron. (b) What magnitude and direction force does this field exert on a proton? Show/Hide Solution Solution(a) $\text{300}\phantom{\rule{0.25em}{0ex}}\text{N/C}\phantom{\rule{0.25em}{0ex}}(\text{east})$ (b) $4\text{.}\text{80}\times {\text{10}}^{\text{17}}\phantom{\rule{0.25em}{0ex}}\text{N}\phantom{\rule{0.25em}{0ex}}(\text{east})$
