Asked by Sameer
three point charges 4nC , 4nC and -3nC are placed at vertices of an equilateral traingle of side 0.20 m. calculate the net electric field at the centriod of the triangle
Answers
Answered by
bobpursley
the centroid of an equilaterial triangle is equidistance from each corner. The distance is
d=(law of sines):
d/sin30=.2/sin120 or d=.1155m
the electric field at the center will be directed towards the negative charge (symettry), so all you have to do is calcualte the portion of each field towards the negative charge.
E=Epos*sin60+Epos*Sin60+Eng
= k(4e-9)*.866*2/.1155^2 + k(-3e-9)/.1155^2=k*5e-9/.1155^2 v/m
d=(law of sines):
d/sin30=.2/sin120 or d=.1155m
the electric field at the center will be directed towards the negative charge (symettry), so all you have to do is calcualte the portion of each field towards the negative charge.
E=Epos*sin60+Epos*Sin60+Eng
= k(4e-9)*.866*2/.1155^2 + k(-3e-9)/.1155^2=k*5e-9/.1155^2 v/m
Answered by
Damon
Hey, they are all 0.2 /sqrt 3 = d from centroid
because cos 30 =.1/d = sqrt 3 /2
so
E from each = k Q/.75
now put them 120 degrees from each other
because cos 30 =.1/d = sqrt 3 /2
so
E from each = k Q/.75
now put them 120 degrees from each other
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