Question

What are the magnitude and direction of the net force on the third spider (S3) when it is resting at the origin

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  magnitude is  $F_3 =0.62\ N$

Explanation:

From the question we are told that

    The distance between the spiders is  $a =0.59\ m$

    The  charge on the $S_1\ and\ S_3$ is  $q_{1}=q_{3}=q =+4.9\mu \ C$

     The  charge on $S_2$ is  $q_2 = 4.9\mu \ C$

Given that the charge are arranged in an equilateral triangle, it implies that the angle at each vertices of the triangle is  60°

Generally the magnitude of net force on $S_3$ is mathematically represented as

   $F_3= F_{13} cos(60)+ F_{21} cos(60)$

So

  $F_{13} =\frac{k* q^2 }{a^2}$

Here k is the columns constant with value   $k = 9*10^9 \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

So

    $F_{13}=\frac{9*10^9*((4.9*10^{-6})^2)}{0.59^2}$

=>  $F_{13} = 0.62 \ N$

From the diagram we see that $F_{13}= F_{23} = 0.62\ N$

So

    $F_3=0.62cos(60)+0.62cos(60)$

     $F_3 =0.62\ N$

Now from the diagram the ls [tex]270^o[\tex] counterclockwise from x-axis