Question

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive x-axis at x = 4.00 cm. (a) If a third charge q3 = +6.00 nC is now placed at the point x = 4.00 cm, y = 3.00 cm, find the x- and y-components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force

Answer 1

Answer:

a

The  x- and y-components of the total force exerted is

           $F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

 The magnitude of the force is  

            $|F_{31 +32}| = 10.25 *10^{-5} N$

   The direction of the force is  

         $\theta =327.43 ^o$   Clockwise from x-axis

Explanation:

From the question we are told that

    The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

      The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

        The position of the second charge  from the first one is  $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

        The  magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

       The position of the third charge from the first one is  $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

                $|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

                $|d_{31}| =5 *10^{-2} m$

        The position of the third charge from the second  one is

                $\= d_{32} = 3j cm = 3j *10^{-2}m$

               $|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

               $|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

           $F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

 Substituting values

          $F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

            $F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

            $F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of     $F_{31 +32}$  is mathematically evaluated as

            $|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

             $|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

            $tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

              $\theta = tan ^{-1} [-0.63889]$

             $\theta = - 32.57 ^o$

             $\theta = 360 - 32.57$

            $\theta =327.43 ^o$