Asked Oct 2, 2019

Three charged particles are located at the corners of an equilateral triangle as shown in the figure below (let q = 3.40 µC, and L = 0.890 m). Calculate the total electric force on the 7.00-µC charge.

Magnitude in Newtons? 

Direction counterclockwise from the x-axis?

7.00 μC
-4.00 μC

Image Transcriptionclose

7.00 μC у + 60.0 -Χ -4.00 μC


Expert Answer

Step 1

Consider the charges on the corners of the equilateral triangle as q, q1, and q2, the distance between the two charged particles as L, the electric force between the particles of charges q and q1 as F1 the electric force between the particles of charges q1 and q2 as F2.


The given values for the equilateral triangle are,


Image Transcriptionclose

q 3.40 uC = 3.40 x 10 C 7.00 C = 7.00 x 106 C q24.00 uC =-4.00 x 10 C L 0.890 m

Step 2

Apply Coulomb’s law to calculate the electric force between the particles of charges q and q1.


Image Transcriptionclose

1 qq 4ле, L (9x10 N m'/C3.40x10 C)(7.00 x 10 C (0.890 m) (3.40x106 C)(7.00 x 10 C) =0.270 N

Step 3

Apply Coulomb’s law to calculate the electric force between the part...


Image Transcriptionclose

1 2 = 4πε, 1 0 7.00 x 10 N m2/C2) C4.00 x 10 C) =(9x10 (0.890 m) --0.318 N


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