What Is the Magnitude of the Electric Field at One Electron Due to the Other?

Learning Objectives

By the cease of this section, yous will exist able to:

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

Contact forces, such every bit between a baseball and a bat, are explained on the pocket-size scale by the interaction of the charges in atoms and molecules in close proximity. They collaborate through forces that include the Coulomb forcefulness. Activity at a distance is a strength between objects that are not close enough for their atoms to "touch." That is, they are separated by more a few diminutive diameters.

For instance, a charged prophylactic comb attracts neutral bits of paper from a altitude via the Coulomb force. It is very useful to remember of an object being surrounded in space by a force field. The force field carries the force to another object (called a examination object) some altitude away.

Concept of a Field

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

In the same fashion, the Coulomb force field surrounding any charge extends throughout space. Using Coulomb'due south constabulary, [latex]F=m\frac{\mid{q_1q_2}\mid}{r^ii}\\[/latex], its magnitude is given by the equation [latex]F=g\frac{\mid{qQ}\mid}{r^ii}\\[/latex], for a point charge (a particle having a charge Q) interim on a examination charge q at a distance r (see Figure 1). Both the magnitude and direction of the Coulomb force field depend on Q and the test charge q.

Effigy one. The Coulomb strength field due to a positive charge Q is shown acting on ii different charges. Both charges are the same distance from Q. (a) Since q ₁ is positive, the force F ₁ acting on it is repulsive. (b) The charge q _two is negative and greater in magnitude than q ₁, and and so the force F _two acting on information technology is attractive and stronger than F ₁. The Coulomb force field is thus not unique at whatsoever point in infinite, considering it depends on the examination charges q _i and q ₂ as well as the charge Q.

To simplify things, we would prefer to have a field that depends only on Q and not on the exam 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 infinite. Specifically, the electric field E is divers to be the ratio of the Coulomb force to the test charge:

[latex]\mathbf{E}=\frac{\mathbf{E}}{q}\\[/latex],

where F is the electrostatic force (or Coulomb force) exerted on a positive test accuse q. It is understood that East is in the aforementioned management equally F. It is too assumed that q is and so pocket-sized that it does not alter the charge distribution creating the electrical field. The units of electric field are newtons per coulomb (N/C). If the electric field is known, and then the electrostatic force on any charge q is only obtained past multiplying charge times electrical field, or F = q E. Consider the electric field due to a point accuse Q. According to Coulomb'south constabulary, the force it exerts on a test charge q is [latex]F=k\frac{\mid{qQ}\mid}{r^2}\\[/latex]. Thus the magnitude of the electrical field, Eastward, for a indicate charge is

[latex]Eastward=\left|\frac{F}{q}\right|=k\left|\frac{qQ}{qr^2}\correct|=thousand\frac{\left|Q\right|}{r^two}\\[/latex].

Since the test charge cancels, nosotros run across that

[latex]E=g\frac{\left|Q\correct|}{r^2}\\[/latex].

The electric field is thus seen to depend merely on the charge Q and the altitude r; information technology is completely independent of the test accuse q.

Example 1. Calculating the Electric Field of a Betoken Accuse

Summate the forcefulness and management of the electrical field E due to a point charge of ii.00 nC (nano-Coulombs) at a altitude of 5.00 mm from the accuse.

Strategy

We can find the electric field created past a betoken charge by using the equation [latex]E=\frac{kQ}{r^ii}\\[/latex].

Solution

Here Q= two.00 × 10^−nine C and r= 5.00 × 10⁻³ m. Entering those values into the to a higher place equation gives

[latex]\begin{array}{lll}E&=&one thousand\frac{Q}{r^two}\\\text{ }&=&\left(8.99\times10^9\frac{\text{N}\cdot\text{m}^two}{\text{C}^2}\right)\times\frac{\left(ii.00\times10^{-9}\text{ C}\right)}{\left(five.00\times10^{-three}\text{ m}\right)^ii}\\\text{ }&=&7.xix\times10^5\text{ N/C}\cease{array}\\[/latex]

Discussion

This electric field force is the same at any point five.00 mm away from the accuse Q that creates the field. It is positive, meaning that it has a management pointing away from the charge Q.

Example 2. Calculating the Force Exerted on a Point Charge by an Electric Field

What forcefulness does the electrical field found in the previous case exert on a point charge of −0.250 μC?

Strategy

Since we know the electric field force and the charge in the field, the strength on that charge can be calculated using the definition of electric field [latex]\mathbf{Eastward}=\frac{\mathbf{F}}{q}\\[/latex] rearranged to F =q Due east.

Solution

The magnitude of the force on a charge q = −0.250 μC exerted by a field of strength Due east = 7.twenty × 10⁵ Due north/C is thus,

[latex]\begin{array}{lll}F&=&-qE\\\text{ }&=&\left(0.250\times10^{-6}\text{ C}\correct)\left(7.20\times10^v\text{ N/C}\correct)\\\text{ }&=&0.180\text{ Due north}\end{array}\\[/latex]

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

Word

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

PhET Explorations: Electrical Field of Dreams

Play ball! Add together charges to the Field of Dreams and see how they react to the electrical field. Plough on a background electric field and adjust the management and magnitude.

Click to download the simulation. Run using Java.

Section Summary

• The electrostatic forcefulness field surrounding a charged object extends out into space in all directions.
• The electrostatic force exerted past a betoken charge on a examination charge at a altitude r depends on the accuse of both charges, as well as the distance betwixt the two.
• The electrical field Eastward is defined to exist [latex]\mathbf{\text{E}}=\frac{\mathbf{\text{F}}}{q}\\[/latex], where F is the Coulomb or electrostatic strength exerted on a small positive test charge q. Eastward has units of Due north/C.
• The magnitude of the electric field E created past a point charge Q is [latex]\mathbf{\text{Eastward}}=k\frac{\left|Q\right|}{{r}^{two}}\\[/latex], where r is the distance from Q. The electric field E is a vector and fields due to multiple charges add like vectors.

Conceptual Questions

1. Why must the examination charge q in the definition of the electric field be vanishingly small-scale?
2. Are the management and magnitude of the Coulomb force unique at a given point in space? What almost the electric field?

Bug & Exercises

1. What is the magnitude and direction of an electric field that exerts a ii.00 × 10⁻⁵ Due north upward force on a −i.75 μC accuse?
2. What is the magnitude and management of the force exerted on a iii.50 μC charge by a 250 N/C electric field that points due e?
3. Calculate 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).
4. (a) What magnitude betoken accuse creates a 10,000 Northward/C electric field at a distance of 0.250 m? (b) How large is the field at ten.0 k?
5. Calculate the initial (from rest) acceleration of a proton in a five.00 × ten⁶ Due north/C electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.
6. (a) Notice the management and magnitude of an electric field that exerts a 4.80 × 10⁻¹⁷ N westward force on an electron. (b) What magnitude and direction forcefulness does this field exert on a proton?

Glossary

field: a map of the amount and direction of a force acting on other objects, extending out into space

indicate charge: A charged particle, designated Q, generating an electric field

exam charge: A particle (designated q) with either a positive or negative charge set downwardly within an electric field generated past a signal charge

Selected Solutions to Problems & Exercises

2. viii.75 × 10^−four N

4. (a) 6.94 × x⁻⁸ C; (b) 6.25 N/C

6. (a) 300 N/C (e); (b) 4.80 × 10⁻¹⁷ N (east)

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Source: https://courses.lumenlearning.com/physics/chapter/18-4-electric-field-concept-of-a-field-revisited/

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