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How to Determine Change in Velocity of a Charged Particle Moved through a Potential Difference
Step 1: Use the formula {eq}\Delta K=qV {/eq} to determine the change in kinetic energy of the particle.
Step 2: Plug the answer from step 1 into the equation {eq}\Delta K=\frac{1}{2}m(v_{f}^2-v_{0}^2) {/eq} and solve for the final velocity of the particle.
Step 3: Find the difference of the initial and final velocities to determine the change in velocity of the particle.
What is a Potential Difference?
Potential Difference: The change in electrical potential, or voltage, a particle experiences at it moves from one location to another location in an electric field. In order for a potential difference to exist there must be an electric field present. If a particle is free to move in a region with a potential difference it will accelerate in a direction parallel to the electric field due to the electric force placed on the particle.
In the case of the particle which is free to accelerate, there is a relationship between the electrical potential through which it is accelerated and the kinetic energy it acquires as it accelerates.
Energy of a Particle Accelerated through a Potential Difference:
$$\Delta U=\Delta K=qV $$
The change in the potential and kinetic energy are equal because there are no outside forces present. Since we can calculate the change in kinetic energy we will also be able to calculate the velocity change.
So, let's try using these steps to determine change in velocity of a charged particle in the following two examples!
Example of How to Determine Change in Velocity of a Charged Particle Moved through a Potential Difference
Example 1
Determine the velocity of an electron accelerated from rest through a potential difference of 50 V. The magnitude of charge on an electron is {eq}1.6\times 10^{-19}\:C {/eq} and the mass of an electron is {eq}9.1\times 10^{-31}\:kg {/eq}.
Step 1: Use the formula {eq}\Delta K=qV {/eq} to determine the change in kinetic energy of the particle.
When plugging into the formula it is not necessary to use the negative sign for the charge of the electron.
$$\Delta K=(1.6\times 10^{-19})(50) $$
$$\Delta K=8.0\times 10^{-18}\:J $$
Step 2: Plug the answer from step 1 into the equation {eq}\Delta K=\frac{1}{2}m(v_{f}^2-v_{0}^2) {/eq} and solve for the final velocity of the particle.
Plugging in the mass, zero for the initial velocity, and the change in kinetic energy gives us
$$8.0\times 10^{-18}=\frac{1}{2}(9.1\times 10^{-31})(v_{f}^2-0^2) $$
Solving for {eq}v_f {/eq} gives us
$$v_f=\sqrt{\frac{2(8.0\times 10^{-18})}{9.1\times 10^{-31}}} $$
$$v_{f}=4.2\times 10^{6}\:m/s $$
Step 3: Find the difference of the initial and final velocities to determine the change in velocity of the particle.
Subtracting zero from the answer to step 2 gives us
$$\Delta v=4.2\times 10^{6}-0=4.2\times 10^{6}\:m/s $$
The change in velocity of the electron is {eq}4.2\times 10^{6}\:m/s {/eq}.
Example 2
Determine the velocity of a proton accelerated from rest through a potential difference of 50 V. The magnitude of the charge on a proton is {eq}1.6\times 10^{-19}\:C {/eq} and the mass of a proton is {eq}1.67\times 10^{-27}\:kg {/eq}.
Step 1: Use the formula {eq}\Delta K=qV {/eq} to determine the change in kinetic energy of the particle.
When plugging into the formula it is not necessary to use the negative sign for the charge of the electron.
$$\Delta K=(1.6\times 10^{-19})(50) $$
$$\Delta K=8.0\times 10^{-18}\:J $$
Step 2: Plug the answer from step 1 into the equation {eq}\Delta K=\frac{1}{2}m(v_{f}^2-v_{0}^2) {/eq} and solve for the final velocity of the particle.
Plugging in the mass, zero for the initial velocity, and the change in kinetic energy gives us
$$8.0\times 10^{-18}=\frac{1}{2}(1.67\times 10^{-27})(v_{f}^2-0^2) $$
Solving for {eq}v_f {/eq} gives us
$$v_f=\sqrt{\frac{2(8.0\times 10^{-18})}{1.67\times 10^{-27}}} $$
$$v_{f}=9.8\times 10^{4}\:m/s $$
Step 3: Find the difference of the initial and final velocities to determine the change in velocity of the particle.
Subtracting zero from the answer to step 2 gives us
$$\Delta v=9.8\times 10^{4}-0=9.8\times 10^{4}\:m/s $$
The change in velocity of the electron is {eq}9.8\times 10^{4}\:m/s {/eq}.
