Three charged particles are at the corners of an equilateral triangle as shown i

Question

Three charged particles are at the corners of an equilateral triangle as shown in the figure below. (Let q = 4.00 mu C, and L = 0.650 m.) (a) Calculate the electric field at the position of charge q due to the 7.00- mu C and -4.00- mu C charges. 106 Once you calculate the magnitude of the field contribution from each charge you need to add these as vectors. kN/C i + -851 Does the -4.00-mu C charge contribute to the y component of the field at the origin? kN/C j (b) Use your answer to part (a) to determine the force on charge q.

Explanation / Answer

a) 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 = (4-7xcos60)x10^-9 x9x10^9 /0.650^2 = 10.65 N/C

b) Once you have the E-field, finding the force is trivial:
F = qE = 4x10^-9x10.65 = 4.26x10^-8 newton

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