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Question: An electron moving in a electric potential field $V_1$ enters a higher electric potential field $V_2...

An electron moving in a electric potential field V1V_1 enters a higher electric potential field V2V_2, then the change in kinetic energy of the electron is proportional to

A

(V2−V1)1/2(V_2-V_1)^{1/2}

B

V2−V1V_2-V_1

C

(V2−V1)2(V_2-V_1)^2

D

(V2V1)V2\frac{(V_2V_1)}{V_2}

Answer

The change in kinetic energy of the electron is proportional to V2−V1V_2-V_1.

Explanation

Solution

The change in kinetic energy of a charged particle moving in an electric field is equal to the work done by the electric field on the particle.

Work done by the electric field on a charge qq moving from potential V1V_1 to V2V_2 is W=q(V1−V2)W = q(V_1 - V_2).

For an electron, q=−eq = -e.

Change in kinetic energy ΔK=W=−e(V1−V2)=e(V2−V1)\Delta K = W = -e(V_1 - V_2) = e(V_2 - V_1).

Thus, ΔK∝(V2−V1)\Delta K \propto (V_2 - V_1).