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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.

[link] shows how the electric field from two point charges can be drawn by finding the total field at representative points and drawing electric field lines consistent with those points. While the electric fields from multiple charges are more complex than those of single charges, some simple features are easily noticed.

For example, the field is weaker between like charges, as shown by the lines being farther apart in that region. (This is because the fields from each charge exert opposing forces on any charge placed between them.) (See [link] and [link] (a).) Furthermore, at a great distance from two like charges, the field becomes identical to the field from a single, larger charge.

[link] (b) shows the electric field of two unlike charges. The field is stronger between the charges. In that region, the fields from each charge are in the same direction, and so their strengths add. The field of two unlike charges is weak at large distances, because the fields of the individual charges are in opposite directions and so their strengths subtract. At very large distances, the field of two unlike charges looks like that of a smaller single charge.

Two charges q one and q two are placed at a distance and their field lines shown by curved arrows move away from each other. At a point P on the field lines emanating from q one, the resultant electric field is represented by a vector arrow tangent to the curve representing this field line. A point P prime on a field line emanating from the charge q two and the resultant electric field is represented by a vector arrow tangent to the curve representing this field line.
Two positive point charges q 1 size 12{q rSub { size 8{1} } } {} and q 2 size 12{q rSub { size 8{2} } } {} produce the resultant electric field shown. The field is calculated at representative points and then smooth field lines drawn following the rules outlined in the text.
In part a, two negative charges of magnitude minus q are placed at some distance. Their field lines are represented by curved arrows terminating into the negative charges. The curves are divergent. In part b, two charges are placed at a distance where one is positive labeled as plus q and other is negative labeled as minus q. The field lines represented by curved arrows start from the positive charge and end at the negative charge. The curves are convergent.
(a) Two negative charges produce the fields shown. It is very similar to the field produced by two positive charges, except that the directions are reversed. The field is clearly weaker between the charges. The individual forces on a test charge in that region are in opposite directions. (b) Two opposite charges produce the field shown, which is stronger in the region between the charges.

Questions & Answers

A soccer player kicked off a ball at velocity of 62 ft/s at angle 45°. A goal keeper is 43 yard away from the direction in which ball kicked off. At what minimum velocity he runs to meet the ball?
Ram Reply
A soccer player kicked off the ball at the velocity of 62 ft/s at 45° with horizontal.A goal keeper is 43 yard away from the ball kicked position.At what minimum velocity he runs to meet the ball?
Ram
what is torque
Deepak Reply
The turning effect of force is called torque.
Uzair
What is the effect of static electricity
Ruth
what there factors affect the surface tension of a liquid
Promise Reply
formula for impedance
muyiwa Reply
ehat is central forces
Nita Reply
what is distance?
Jonathan Reply
What does mean ohms law imply
Victoria Reply
ohms law state that the electricity passing through a metallic conductor is directly proportional to the potential difference across its end
muyiwa
what is matter
folajin Reply
Anything that occupies space
Kevin
Any thing that has weight and occupies space
Victoria
Anything which we can feel by any of our 5 sense organs
Suraj
Right
Roben
thanks
Suraj
what is a sulphate
Alo
any answers
Alo
the time rate of increase in velocity is called
Blessing Reply
acceleration
Emma
What is uniform velocity
Victoria
Greetings,users of that wonderful app.
Frank Reply
how to solve pressure?
Cruz Reply
how do we calculate weight and eara eg an elefant that weight 2000kg has four fits or legs search of surface eara is 0.1m2(1metre square) incontact with the ground=10m2(g =10m2)
Cruz
P=F/A
Mira
can someone derive the formula a little bit deeper?
Bern
what is coplanar force?
OLADITI Reply
forces acting and lying on d same plane
Promise
what is accuracy and precision
Peace Reply
How does a current follow?
Vineeta Reply
follow?
akif
which one dc or ac current.
akif
how does a current following?
Vineeta
?
akif
AC current
Vineeta
AC current follows due to changing electric field and magnetic field.
akif
you guys are just saying follow is flow not follow please
Abubakar
ok bro thanks
akif
flows
Abubakar
but i wanted to understand him/her in his own language
akif
but I think the statement is written in English not any other language
Abubakar
my mean that in which form he/she written this,will understand better in this form, i write.
akif
ok
Abubakar
ok thanks bro. my mistake
Vineeta
u are welcome
Abubakar
what is a semiconductor
Vineeta Reply
substances having lower forbidden gap between valence band and conduction band
akif
what is a conductor?
Vineeta
replace lower by higher only
akif
convert 56°c to kelvin
Abubakar
How does a current follow?
Vineeta
A semiconductor is any material whose conduction lies between that of a conductor and an insulator.
AKOWUAH
Practice Key Terms 4

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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