Question

2. Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. The charges are +7.0μC, -8.0 μC and -6.0 μC. Calculate the net force on charge 1 due to the other two charges in unit vector notation. Give values for the magnitude and direction of the force, too.

Answer 1

Answer:

0.53 N, 25.6°

Explanation:

side of triangle, a = 1.2 m

q = 7 μC

q1 = - 8 μC

q2 = - 6 μC

Let F1 be the force between q and q1

By using the coulomb's law

$F_{1}=\frac{Kq_{1}q}{a^{2}}$

$F_{1}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 8\times 10^{-6}}{1.2^{2}}$

F1 = 0.35 N

Let F2 be the force between q and q2

By using the coulomb's law

$F_{2}=\frac{Kq_{2}q}{a^{2}}$

$F_{2}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 6\times 10^{-6}}{1.2^{2}}$

F2 = 0.26 N

Write the forces in the vector form

$\overrightarrow{F_{1}}=0.35\widehat{i}$

$\overrightarrow{F_{2}}=0.26\left ( Cos60 \widehat{i}+Sin60\widehat{j}\right )$

$\overrightarrow{F_{2}}=0.13 \widehat{i}+0.23\widehat{j}$

Net force

$\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}$

$\overrightarrow{F}=0.48 \widehat{i}+0.23\widehat{j}$

Magnitude of the force

$F=\sqrt{0.48^{2}+0.23^{2}}$

F = 0.53 N

Direction of force with x axis

$tan\theta =\frac{0.23}{0.48}$

θ = 25.6°