Question

Three charged particles are at the corners of an equilateral
triangle as shown in the figure below. (Let *q* = 4.00

Answer #1

a) You have to break the E-fields into components.

Let q1 be 4 microcoulombs and q2 be 7 microcoulombs. The variables will be positive. I'll deal with the negative charge on q1 "by hand".

E-field1 = k (q1) / L^2.

Because q1 is negative, the E-field's direction at q is to the
right (positive i-hat), toward q1.

E-field2 = k(q2) / L^2

Because q2 is positive, the field direction at q is left and down,
away from q2.

(-cos(60) i-hat - sin(60) j-hat)

Multiply that out.

You've got the vertical component: - sin (60) k(q2) / L^2

You have to subtract the horizontal component of the second E-field from the first: (q1 - q2 cos(60)) k / L^2

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