Question

An electron moving parallel to the x axis has an initial speed of 4.10 106 m/s at the origin. Its speed is reduced to 1.76 105 m/s at the point x = 2.00 cm.
Required:
a. Calculate the electric potential difference between the origin and that point.
b. Which point is at the higher potential?

Answer 1

Answer:

a

  $V = -47.65 N/C$

b

At the origin

Explanation:

From the question we are told that

   The  initial speed is  $v_1 = 4.10 *10^{6} \ m/s$

    The  speed at (x = 2.00 cm) is  $v_f = 1.76 *10^{5} \ m/s$

   

Generally the electric potential difference is mathematically represented as

    $V = \frac{W}{q}$

Here W is the  work-done which is mathematically represented as

    $W = K_f - K_i$

Here  $K_f$ is the kinetic energy at  x =  2.00 cm mathematically expressed as

        $K_f = \frac{1}{2} * m* v^2_f$

and  

       $K_i$ is the kinetic energy at origin  mathematically expressed as

        $K_f = \frac{1}{2} * m* v^2_i$

So

    $V = \frac{1}{q} [ \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2]$

    $V = \frac{m}{2q} [v_f^2 - v_i^2]$

Here m is the mass of electron with value $m = 9.1*10^{-31} \ kg$

         q  is the charge on the electron with value  $q = 1.60*10^{-19} \ C$

So

     $V = \frac{(9.1 *10^{-31})}{2(1.60*10^{-19})} [(1.76*10^{5})^2 - (4.10*10^{6})^2]$

     $V = -47.65 N/C$

So given that the difference is negative then potential is higher at the origin