# How to calculate the maximum velocity of a charged particle in an electric field?

Explained simply in a way I could use to answer and exam questions and understand the concept, please.

Kia ora Sophie,

Great question. Let’s start with a concept you might be familiar with from Level 1. When we drop an object, an unbalanced force causes it to accelerate towards the ground. As it falls it is moving through Earth’s gravitational field. Potential energy (Ep) is converted into kinetic energy (Ek). The amount of Ep lost is the same as the amount of Ek gained. If you did L1 Mechanics, you may recall using Ep=mgh and Ek=1/2mv^2 to work out how much Ep an object has, then working out it’s velocity based on gaining that amount of kinetic energy.

Electric fields are very similar. Inside an electric field a charged particle will experience an unbalanced force causing it to accelerate. (Eg a negatively charged electron would be repelled by the negative plate and attracted by the positive plate. As the electron accelerates it’s velocity increases). We can work out the lost potential energy of the electron with Ep=Eqd, where:
E is the electric field strength
q is the charge of the electron (1.6x10^-19 Coulombs - this is given in your exam resource booklet)
d is the distance the electron moves (in metres)

The Ep that is lost, is converted into kinetic energy (Ek). We can rearrange Ek=1/2mv^2 into v=√ (2Ek/m) to find the velocity of the electron (mass = 9.11x10^-31 kg, this is given in the resource booklet).

No Brain Too Small has collated these questions, along with answers. The 2016 question might be a good one to start with.

Once you feel you understand that, you could look at an example where the velocity is given, but you have to use the same equations to calculate the ditance between the ‘plates’.

I hope that helps. If you’re still struggling with the concept, let me know and I’ll put together a video example for you.