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Note that the electric field is defined for a positive test charge q size 12{q} {} , so that the field lines point away from a positive charge and toward a negative charge. (See [link] .) The electric field strength is exactly proportional to the number of field lines per unit area, since the magnitude of the electric field for a point charge is E = k | Q | / r 2 size 12{E= { ital "kQ"} slash {r rSup { size 8{2} } } } {} and area is proportional to r 2 size 12{r rSup { size 8{2} } } {} . This pictorial representation, in which field lines represent the direction and their closeness (that is, their areal density or the number of lines crossing a unit area) represents strength, is used for all fields: electrostatic, gravitational, magnetic, and others.

In part a, electric field lines emanating from a positive charge is shown by the vector arrows in all direction of two dimensional space and the density of these field lines is less. In part b, electric field lines entering the negative charge is shown by the vector arrows coming from all direction of two dimensional space and the density of these field lines is less. In part c, electric field lines entering the negative charge is shown by the vector arrows coming from all direction of two dimensional space and the density of these field lines is large.
The electric field surrounding three different point charges. (a) A positive charge. (b) A negative charge of equal magnitude. (c) A larger negative charge.

In many situations, there are multiple charges. The total electric field created by multiple charges is the vector sum of the individual fields created by each charge. The following example shows how to add electric field vectors.

Adding electric fields

Find the magnitude and direction of the total electric field due to the two point charges, q 1 size 12{q rSub { size 8{1} } } {} and q 2 size 12{q rSub { size 8{2} } } {} , at the origin of the coordinate system as shown in [link] .

Two charges are placed on a coordinate axes. Q two is at the position x equals 4 and y equals 0 centimeters. Q one is at the position x equals 0 and y equals two centimeters. Charge on q one is plus five point zero nano coulomb and charge on q two is plus ten nano coulomb. The electric field, E one having a magnitude of one point one three multiplied by ten raise to the power five Newton per coulomb is represented by a vector arrow along positive y axis starting from the origin. The electric field, E two having magnitude zero point five six multiplied by ten raise to the power five Newton per coulomb is represented by a vector arrow along negative x axis starting from the origin. The resultant field makes an angle of sixty three point four degree above the negative y axis having magnitude one point two six multiplied by ten raise to the power five Newton per coulomb is represented by a vector arrow pointing away from the origin in the second quadrant.
The electric fields E 1 size 12{E rSub { size 8{1} } } {} and E 2 size 12{E rSub { size 8{2} } } {} at the origin O add to E tot size 12{E rSub { size 8{"tot"} } } {} .

Strategy

Since the electric field is a vector (having magnitude and direction), we add electric fields with the same vector techniques used for other types of vectors. We first must find the electric field due to each charge at the point of interest, which is the origin of the coordinate system (O) in this instance. We pretend that there is a positive test charge, q size 12{q} {} , at point O, which allows us to determine the direction of the fields E 1 size 12{E rSub { size 8{1} } } {} and E 2 size 12{E rSub { size 8{2} } } {} . Once those fields are found, the total field can be determined using vector addition    .

Solution

The electric field strength at the origin due to q 1 size 12{q rSub { size 8{1} } } {} is labeled E 1 size 12{E rSub { size 8{1} } } {} and is calculated:

E 1 = k q 1 r 1 2 = 8 . 99 × 10 9 N m 2 /C 2 5 . 00 × 10 9 C 2 . 00 × 10 2 m 2 E 1 = 1 . 124 × 10 5 N/C . alignl { stack { size 12{E rSub { size 8{1} } =k { {q rSub { size 8{1} } } over {r rSub { size 8{1} } rSup { size 8{2} } } } = left (9 "." "00" times "10" rSup { size 8{9} } N cdot m rSup { size 8{2} } "/C" rSup { size 8{2} } right ) { { left (5 "." "00" times "10" rSup { size 8{ - 9} } C right )} over { left (2 "." "00" times "10" rSup { size 8{ - 2} } m right ) rSup { size 8{2} } } } } {} #E rSub { size 8{1} } =1 "." "125" times "10" rSup { size 8{5} } "N/C" {} } } {}

Similarly, E 2 size 12{E rSub { size 8{2} } } {} is

E 2 = k q 2 r 2 2 = 8 . 99 × 10 9 N m 2 /C 2 10 . 0 × 10 9 C 4 . 00 × 10 2 m 2 E 2 = 0 . 5619 × 10 5 N/C . alignl { stack { size 12{E rSub { size 8{2} } =k { {q rSub { size 8{2} } } over {r rSub { size 8{2} } rSup { size 8{2} } } } = left (9 "." "00" times "10" rSup { size 8{9} } N cdot m rSup { size 8{2} } "/C" rSup { size 8{2} } right ) { { left ("10" "." 0 times "10" rSup { size 8{ - 9} } C right )} over { left (4 "." "00" times "10" rSup { size 8{ - 2} } m right ) rSup { size 8{2} } } } } {} #E rSub { size 8{2} } =0 "." "5625" times "10" rSup { size 8{5} } "N/C" {} } } {}

Four digits have been retained in this solution to illustrate that E 1 size 12{E rSub { size 8{1} } } {} is exactly twice the magnitude of E 2 size 12{E rSub { size 8{2} } } {} . Now arrows are drawn to represent the magnitudes and directions of E 1 size 12{E rSub { size 8{1} } } {} and E 2 size 12{E rSub { size 8{2} } } {} . (See [link] .) The direction of the electric field is that of the force on a positive charge so both arrows point directly away from the positive charges that create them. The arrow for E 1 size 12{E rSub { size 8{1} } } {} is exactly twice the length of that for E 2 size 12{E rSub { size 8{2} } } {} . The arrows form a right triangle in this case and can be added using the Pythagorean theorem. The magnitude of the total field E tot size 12{E rSub { size 8{"tot"} } } {} is

E tot = ( E 1 2 + E 2 2 ) 1/2 = { ( 1.124 × 10 5 N/C ) 2 + ( 0.5619 × 10 5 N/C ) 2 } 1/2 = 1.26 × 10 5 N/C. alignl { stack { size 12{E rSub { size 8{ ital "tot"} } `= \( E rSub { size 8{1} } rSup { size 8{2} } `+`E rSub { size 8{2} } rSup { size 8{2} } \) rSup { size 8{ {1} wideslash {2} } } } {} #~``=` lbrace \( 1 "." "125" times "10" rSup { size 8{5} } `"N/C" \) rSup { size 8{2} } `+` \( 0 "." "5625" times "10" rSup { size 8{5} } `"N/C" \) rSup { size 8{2} } rbrace rSup { size 8{ {1} wideslash {2} } } {} # `~`=``1 "." "26" times "10" rSup { size 8{5} } `"N/C" {}} } {}

The direction is

θ = tan 1 E 1 E 2 = tan 1 1 . 124 × 10 5 N/C 0 . 5619 × 10 5 N/C = 63 . , alignl { stack { size 12{θ="tan" rSup { size 8{ - 1} } left ( { {E rSub { size 8{1} } } over {E rSub { size 8{2} } } } right )} {} #="tan" rSup { size 8{ - 1} } left lbrace { {1 "." "125" times "10" rSup { size 8{5} } " N/C"} over {0 "." "5625" times "10" rSup { size 8{5} } " N/C"} } right rbrace {} # ="63" "." 4° {}} } {}

or 63.4º above the x -axis.

Discussion

In cases where the electric field vectors to be added are not perpendicular, vector components or graphical techniques can be used. The total electric field found in this example is the total electric field at only one point in space. To find the total electric field due to these two charges over an entire region, the same technique must be repeated for each point in the region. This impossibly lengthy task (there are an infinite number of points in space) can be avoided by calculating the total field at representative points and using some of the unifying features noted next.

Questions & Answers

an 8.0 capacitor is connected by to the terminals of 60Hz whoes rms voltage is 150v. a.find the capacity reactance and rms to the circuit
Aisha Reply
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Valdes Reply
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Nwafor Reply
is the study of matter in relation to energy
Kintu
a submersible pump is dropped a borehole and hits the level of water at the bottom of the borehole 5 seconds later.determine the level of water in the borehole
Obrian Reply
what is power?
aron Reply
power P = Work done per second W/ t. It means the more power, the stronger machine
Sphere
e.g. heart Uses 2 W per beat.
Rohit
A spherica, concave shaving mirror has a radius of curvature of 32 cm .what is the magnification of a persons face. when it is 12cm to the left of the vertex of the mirror
Alona Reply
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Shii
1.75cm
Ridwan
my name is Abu m.konnek I am a student of a electrical engineer and I want you to help me
Abu
the magnification k = f/(f-d) with focus f = R/2 =16 cm; d =12 cm k = 16/4 =4
Sphere
A weather vane is some sort of directional arrow parallel to the ground that may rotate freely in a horizontal plane. A typical weather vane has a large cross-sectional area perpendicular to the direction the arrow is pointing, like a “One Way” street sign. The purpose of the weather vane is to indicate the direction of the wind. As wind blows pa
Kavita Reply
hi
Godfred
what about the wind vane
Godfred
If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
Anurag Reply
the same behavior thru the prism out or in water bud abbot
Ju
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
Anurag
What was the previous far point of a patient who had laser correction that reduced the power of her eye by 7.00 D, producing a normal distant vision power of 50.0 D for her?
Jaydie Reply
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Jaydie
29/20 ? maybes
Ju
In what ways does physics affect the society both positively or negatively
Princewill Reply
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rerry Reply
try to read several books on phy don't just rely one. some authors explain better than other.
Ju
And don't forget to check out YouTube videos on the subject. Videos offer a different visual way to learn easier.
Ju
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Ju
I have a exam on 12 february
David Reply
what is velocity
Jiti
the speed of something in a given direction.
Ju
what is a magnitude in physics
Jiti Reply
Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
Giovani Reply
What is meant by dielectric charge?
It's Reply
Practice Key Terms 4

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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