Question

An electron is accelerated through a potential difference of 0.20 v. How much greater would its final speed be if it is accelerated with four times as much voltage?

Answer 1

Answer:

v₂ = 2* v₁

Explanation:

• Assuming no friction present, the change in the electrostatic potential energy of the electron, must be equal in magnitude to the change in the kinetic energy:

       $\Delta U_{ep} = (-e)*\Delta V= - \Delta K = -\frac{1}{2}*m*v^{2}$

• We know that ΔV = 0.2 V, so we can write the following equality:

        $e* 0.2V = \frac{1}{2} * m *v_{1} ^{2} (1)$

• Now, if the voltage increases 4 times, we can write the following equality:

       $e* 0.8V = \frac{1}{2} * m *v_{2} ^{2} (2)$

• Dividing both sides in (1) and (2), and simplifying common terms, we get:

        $\frac{v_{2} ^{2}}{v_{1} ^{2}} = 4\\\\ v_{2} ^{2} = 4*v_{1} ^{2}$

• Taking square roots at both sides, we get:
• v₂ = 2* v₁
• The final speed, if the electron is accelerated through a potential difference four times larger, will be double than the one for a potential difference of 0.2 V.