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ACHE
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THREE CHARGES ARE AT THE corners of equilateral triangle.calculate electric feild at left corner charge?
 2 years ago
 2 years ago
ACHE Group Title
THREE CHARGES ARE AT THE corners of equilateral triangle.calculate electric feild at left corner charge?
 2 years ago
 2 years ago

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henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1349626330386:dw
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1349626433674:dw
 2 years ago

henpen Group TitleBest 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
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[E=\frac{F}{q}=\frac{Qk}{r^2}\]
 2 years ago

ACHE Group TitleBest ResponseYou've already chosen the best response.0
i dnt understand:/
 2 years ago

henpen Group TitleBest 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.
 2 years ago

ACHE Group TitleBest ResponseYou've already chosen the best response.0
utube iz bloked now:/
 2 years ago

ACHE Group TitleBest 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.
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1349627494021:dw Are you fine this far?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
http://gyazo.com/01feb6fa59915fa730dc10d234ac6a87
 2 years ago

henpen Group TitleBest 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=\[\frac{7* (9*10^9)}{r^2} sin(60)\]
 2 years ago

henpen Group TitleBest 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!
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
*Multiply E field by the charge to get force
 2 years ago

ACHE Group TitleBest ResponseYou've already chosen the best response.0
dw:1349627940207:dw
 2 years ago

ACHE Group TitleBest ResponseYou've already chosen the best response.0
r sq is o.5 sq. na?
 2 years ago
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