Question

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Answer 1

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

   The  charge on the first sphere is  $q_1 = 2\mu C = 2*10^{-6} \ C$

    The charge on the second sphere is  $q_2 = 8 \mu C = 8*10^{-6} \ C$

     The  mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

      The  distance apart is  $d = 0.4 \ m$

      The  speed of the second  sphere is  $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and  $q_1$ are separated by $0.8 \ m$ is mathematically represented

     $Q = KE + U$

Here KE   is  the kinetic energy which is mathematically represented as

     $KE = \frac{1 }{2} m (v_1)^2$

substituting value

     $KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

     $KE = 0.3 \ J$

And  U is  the  potential  energy which is mathematically represented as

        $U = \frac{k * q_1 * q_2 }{d }$

substituting values

       $U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

      $U = 0.18 \ J$

So

       $Q = 0.3 + 0.18$

       $Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and  $q_1$ are separated by $0.4 \ m$ is mathematically represented

         $Q_f = KE_f + U_f$

Here $KE_f$ is  the kinetic energy which is mathematically represented as

     $KE_f = \frac{1 }{2} m (v_2^2$

substituting value

     $KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

     $KE_f = 7.50 *10^{ -4} (v_2 )^2$

And  $U_f$ is  the  potential  energy which is mathematically represented as

        $U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

       $U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

      $U_f = 0.36 \ J$

From the law of energy conservation

     $Q = Q_f$

So

    $0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

   $v_2 = 4 \sqrt{10} \ m/s$