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

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henpen
 2 years ago
Best ResponseYou've already chosen the best response.1How 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

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

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

ACHE
 2 years ago
Best ResponseYou've already chosen the best response.0theree 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.

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

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

henpen
 2 years ago
Best 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!

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