Question

An electron that is initially 52 cm away from a proton is displaced to another point. If the change in the electrical potential energy as a result of this movement is 2.4 × 10−28 J, what is the final distance between the electron and the proton? The Coulomb constant is 8.98755 × 109 N · m2 /C 2 , the acceleration due to gravity is 9.8 m/s 2 and the fundamental charge is 1.602 × 10−19 C . Answer in units of m.

Answer 1

Explanation:

The given data is as follows.

       Change in Potential Energy = $2.4 \times 10^{-28} J$

Formula for change in potential energy is as follows

       Change in Potential Energy = $charge \times {\text{potential difference}}$

Hence, the potential difference between the two points will be as follows.

               V = $\frac{2.4 \times 10^{-28} J}{-1.602 \times 10^{-19}}$

                   = $-1.5 \times 10^{-9}$ volt

Therefore, the potential due to two charge initially is as follows.

                     V = $V_{f} - V_{i}$

                         = $-1.5 \times 10^{-9}$ volt

Hence,    $-1.5 \times 10^{-9}$ volt = $9 \times 10^{9} \times 1.602 \times 10^{-19}[\frac{1}{r} - \frac{1}{0.52}]$

                                  r = 0.493 m = 49.3 cm    (as 1 m = 100 cm)

Thus, we can conclude that the final distance between the electron and proton is 49.3 cm.