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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 body travelling at a velocity of 30ms^-1 in a straight line is brought to rest by application of brakes. if it covers a distance of 100m during this period, find the retardation.
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The change in position of an object with respect to time
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Pamilerin
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Stephen
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Laura
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Laura
acceleration it is the rate of change in velocity with time
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acceleration is change in velocity per rate of time
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mass × acceleration OR Work done ÷ distance
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because of gravitational forces
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E=MC^2
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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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