Question

Three charged particles are located at the corners of an equilateral triangle as shown in the figure below (let q=3.20 μC, and L = 0.870 m). Calculate the total electric force on the 7.00-μC charge.

Answer 1

knowing that

\(F_{13}=0.26 N\)

where \(q_{1}=3.20 \mu C=3.20 x 10^{-6} \mathrm{C}\) and \(q_{3}=7 \mu C=7 \times 10^{-6} \mathrm{C}\)

knowing that

\(F_{23}=0.33 N\)

where \(q_{2}=-4 \mu C=-4 x 10^{-6} C\)

Note: the amounts can vary by the decimals considered.

Using the equation to determine the components of a vector, we obtain:

$$ F_{13 x}=F_{13} \cdot \cos 60^{0}=(0.26 N) \cos 60^{0}=0.13 N $$

\(F_{13 y}=F_{13} \cdot \sin 60^{0}=(0.26 N) \sin 60^{\circ}=0.23 N\)

\(F_{23 x}=F_{23} \cdot \cos 60^{0}=(0.33 N) \cos 60^{0}=0.165 N\)

\(F_{23 y}=F_{23} \cdot \sin 60^{0}=(0.33 N) \sin 60^{\circ}=0.29 N\)

Now we determine the component in \(x\) and the component in \(y\)

$$ F_{3 x}=F_{13 x}+F_{23 x}=0.13 N+0.165 N=0.295 N $$

\(F_{3 y}=F_{13 y}+F_{23 y}=0.23 N+0.29 N=0.52 N\)

Finally the direction is:

\(F_{3}=(0.295 \hat{i}+0.52 \hat{j}) N\)

Note: the amounts can vary by the decimals considered.