Question

An electron at an initial electrical potential of 0 V is fired towards a second electron which is held fixed in space. The moving electron was fired at an initial speed of 1 x 10^4 m/s. When the moving electron is 5.11 x 10^-6 m from the fixed electron it's speed has been reduced to 1 x 10^3 m/s. What is the electrical potential 5.11 x 10^-6 m at the point 5.11 x 10^-6 m away from an isolated electron?

Answer 1

Here , let the potential at this point is Vf

Now , Using conservation of energy

initial mechanical energy = final mechanical energy

0.50 * 9.11 *10^-31 * (1 *10^4)^2 - 1.602 *10^-19 * 0 = 0.50 * 9.11 *10^-31 * (1 *10^3)^2 - 1.602 *10^-19 * Vf

solving for Vf

Vf = - 0.000281 V

the potential at this point is - 0.000281 V