Question

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
Required:

What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge?

Answer 1

Complete Question

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

Required:

What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?

Answer:

The velocity is $v = 80.82 \ m/s$

Explanation:

From  the question we are told that

   The magnitude of charge one is  $q_1 = 3.25 nC = 3.25 *10^{-9} \ C$

   The  magnitude of charge two  $q_2 = 2.00 \ nC = 2.00 *10^{-9} \ C$

   The distance of separation is   $d = 58.0 \ cm = 0.58 \ m$

 

Generally the electric potential of the electron at the midway point is mathematically represented as

         $V = \frac{ q_1 }{\frac{d}{2} } + \frac{ q_2}{\frac{d}{2} }$

substituting values

         $V = \frac{ 3.25 *10^{-9} }{\frac{ 0.58}{2} } + \frac{ 2 *10^{-9} }{\frac{ 0.58}{2} }$

          $V = 1.8103 *10^{-8} \ V$

Now when the electron is 10 cm   =  0.10 m  from charge 1 , it is  (0.58 - 0.10 =  0.48 m ) m from charge two

Now the electric potential  at that point is mathematically represented as

       $V_1 = \frac{q_1}{ 0.10} + \frac{q_2}{ 0.48}$

 substituting values

      $V_1 = \frac{3.25 *10^{-9}}{ 0.10} + \frac{2.0*10^{-9}}{ 0.48}$

      $V_1 = 3.67*10^{-8} \ V$

Now the law of energy conservation ,

   The  kinetic energy of the electron  =  potential energy of the electron

i.e     $\frac{1}{2} * m * v^2 = [V_1 - V]* q$

where q is the magnitude of the charge on the electron with value

      $q = 1.60 *10^{-19} \ C$

While m is the mass of the electron with value  $m = 9.11*10^{-31} \ kg$

        $\frac{1}{2} * 9.11 *10^{-19} * v^2 = [ (3.67 - 1.8103) *10^{-8}]* 1.60 *10^{-19}$

         $v = \sqrt{6532.4}$

        $v = 80.82 \ m/s$