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THREE CHARGES ARE AT THE corners of equilateral triangle.calculate electric feild at left corner charge?
 one year ago
 one year ago
THREE CHARGES ARE AT THE corners of equilateral triangle.calculate electric feild at left corner charge?
 one year ago
 one year ago

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henpenBest ResponseYou've already chosen the best response.1
How would you actually do this? Well, define E2 to be on the xaxis, so \[E_{2(x)}=E_2\]\[E_{2(y)}=0\] And\[E_{1(x)}=E_1\cos(60)\]\[E_{1(y)}=E_1\sin(60)\]Add the x and y components
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[E=\frac{F}{q}=\frac{Qk}{r^2}\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
http://www.youtube.com/watch?v=OsWDUqJQcpk Watch the first 9 minutes, he explains something very similar to your problem on the 7th9th minutes.
 one year ago

ACHEBest ResponseYou've already chosen the best response.0
theree charges are at the corners of an equilateral triangle as shown in fig below , calculte the electric feild at the position of tha 2.00 micro colum charge due to the 7.00 micro columb and 4.00 micro columb charges.(b) use ur answer to detrmine the force on the 2.00 mdw:1349627210442:dwicro columb chare.
 one year ago

henpenBest ResponseYou've already chosen the best response.1
dw:1349627494021:dw Are you fine this far?
 one year ago

henpenBest ResponseYou've already chosen the best response.1
http://gyazo.com/01feb6fa59915fa730dc10d234ac6a87
 one year ago

henpenBest ResponseYou've already chosen the best response.1
+ve direction=right E1x=\[\frac{4* (9*10^9)}{r^2}\] E2x=\[\frac{7* (9*10^9)}{r^2} cos(60)\] E2y=\[\frac{7* (9*10^9)}{r^2} sin(60)\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[E=F/q\]Multiply by the charge Sorry! In my earlier calculations it should have been MICRO coulombs multiply each by 10^6!
 one year ago

henpenBest ResponseYou've already chosen the best response.1
*Multiply E field by the charge to get force
 one year ago
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