Question

In the figure, the particles have charges q1 =-q2-600 nC and q,--q,-98 nC, and distance a = 4.8 cm. what are the (a) x and (b) y components of the net electrostatic force on particle 3? (a) Number0.119 Units (b) Number0. 148 Units

Answer 1

(a)

$F_x=\frac {kq_3^2}{a^2}+\frac {kq_2q_3}{2a^2}\cos 45^{\degree}=\frac {k}{2a^2}(2q_3^2+q_2q_3\cos 45^{\degree})$

$F_x=\frac {9\cdot 10^9}{2\cdot (4.8\cdot 10^{-2})^2}(2\cdot (600\cdot 10^{-9})^2+98\cdot 600\cdot 10^{-18}\cdot 0.707)$

$F_x=1.487\: N$

(b)

$F_{31}=-\frac {kq_1q_3}{a^2}=\frac {9\cdot 10^9\cdot 600\cdot 10^{-9}\cdot 98\cdot 10^{-9}}{0.048^2}=-0.229\: N$

$F_{32}=\frac {kq_2q_3}{2a^2}\cos 45^{\degree}=\frac {9\cdot 10^9\cdot 600\cdot 10^{-9}\cdot 98\cdot 10^{-9}\cos 45^{\degree}}{2\cdot 0.048^2}=0.0812\: N$

$F_y=F_{31}+F_{32}=-0.1478\: N$