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Myr73
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Consider an equilateral triangle of side 15.6 cm. A charge of +2.0uc is placed at one vertex and charges of -4.0C uc each are placed at the other two, as shown in the diagram to the right. Determine the electric field at the centre of the triangle
ANgle= 60 sides--> d1= d2=d3=0.156m q1=2X10^-6C q2=q3=-4X10^-6C
E(center)=?
E=kQ/r^2 {E= E1+E2+E3} Ex= Ex1 +Ex2+Ex3 Ey= Ey1+Ey2+Ey3
Ex.=(+-)E.CosAngle Ey.=(+-)E.Sin(angle)
Let q1 be the charge on top, q2 bottom left, q3 bottom right. Side d1, left side, side d2 , right side and side d3, bottom side of triangle.
I am having trouble witch direction the E components are going. That is for example, which direction is the Ex2 going. For example if its going towards the left then I would put a negative sign in front of E2 --> Ex2=-E2Cos(angle)
Is it the following ? that E2x and E2y would be negative,(pointing out of the triangle) E3x would be positive,E3y would be negative,(also pointing out,towards charge 3) and E1x is N/A, and E1y is negative(poiting towards the middle )?
ANgle= 60 sides--> d1= d2=d3=0.156m q1=2X10^-6C q2=q3=-4X10^-6C
E(center)=?
E=kQ/r^2 {E= E1+E2+E3} Ex= Ex1 +Ex2+Ex3 Ey= Ey1+Ey2+Ey3
Ex.=(+-)E.CosAngle Ey.=(+-)E.Sin(angle)
Let q1 be the charge on top, q2 bottom left, q3 bottom right. Side d1, left side, side d2 , right side and side d3, bottom side of triangle.
The Attempt at a Solution
I am having trouble witch direction the E components are going. That is for example, which direction is the Ex2 going. For example if its going towards the left then I would put a negative sign in front of E2 --> Ex2=-E2Cos(angle)
Is it the following ? that E2x and E2y would be negative,(pointing out of the triangle) E3x would be positive,E3y would be negative,(also pointing out,towards charge 3) and E1x is N/A, and E1y is negative(poiting towards the middle )?