Positive point charges of 10 x 10^-3 C are places at two vertices of an equilate
ID: 2170411 • Letter: P
Question
Positive point charges of 10 x 10^-3 C are places at two vertices of an equilateral triangle with sides 2.0 m. calculate the magnitude of the electric field at the third vertex.My notes: It is an equilateral triangle
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2.0m / 2.0 meter
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2.0 meter
I know the formula is E= (K(Q))/r^2
It says calculate the third vertex so im not sure if i have to find each vertex or just plug in the numbers into the formula. if its only plugging in the numbers than im pretty sure its {(9.0*10^-6)x(10*10^-6)}/2.0^2 can someone make sure if im right and if im wrong can someone help me.
Explanation / Answer
The field at the vertex from each point is equal in magnitude E1 = k*q/r^2 = 9.0x10^9*10 x 10^-3/2m = N/C The net field will be along the bisector of the angle and equals E1*cos(30) So the total field is 2*E1*cos(30) = 2*E1*cos(30) = N/C Right I didn't square in my calculations So E1 = k*q/r^2 = 9.0x10^9*10 x 10^-3 C/(2)^2= N/C So the total field is 2*E1*cos(30) = 2*E1/C*cos(30) = N/C