Electric fields help!

Can you please explain the answer to the following to me:

A positive ion has a charge to mass ratio of 2.40*10^7 Ckg^-1. It is held stationary in a vertical electric field.

What is the electric field strength and its direction?

Many thanks!

The electric filed strength is force per unit charge, $E=\dfrac{F}{Q}$ and you know that if the potential difference between the plates was zero, the charge would fall under gravity as a result of its weight. Therefore, since the charge is held between the plates, you should realise that the charge must be attracted towards the "upper plate" (which must be the negative plate as the charge is positive) and that the electric force on the charge must be equal to the weight of the charge. Hence, you can write $E=\dfrac{F}{Q}=\dfrac{mg}{Q}= \dfrac{m}{Q}g$. I'm sure you can now work out the direction of the electric field.

The electric filed strength is force per unit charge, $E=\dfrac{F}{Q}$ and you know that if the potential difference between the plates was zero, the charge would fall under gravity as a result of its weight. Therefore, since the charge is held between the plates, you should realise that the charge must be attracted towards the "upper plate" (which must be the negative plate as the charge is positive) and that the electric force on the charge must be equal to the weight of the charge. Hence, you can write $E=\dfrac{F}{Q}=\dfrac{mg}{Q}= \dfrac{m}{Q}g$. I'm sure you can now work out the direction of the electric field.

Also i am probably being ridiculously dim but how do i know for a fact that the potential difference between the plates are zero ? Is that a fact i just need to know because it doesn’t state in the question ?