# Electric fields help!

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!
Sorry you've not had any responses about this. Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses.

I'm going to quote in Tank Girl now so she can move your thread to the right place if it's needed.

Spoiler

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.
I know this was ages ago but I just found this and it really helped me so thanks!
Original post by Absent Agent
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.
Just been practicing my multiple choice for my exam on Friday and got stuck on this for ages, what a legend you are
Original post by Absent Agent
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.
Hey i know this thread is old but can someone pls explain to me how i know the direction of the electric field strength ?
(edited 1 year ago)
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 ?
There would be a potential difference, as it is said it is 'held stationary' in the magnetic field. There is no resultant force acting. on the positive ion in order for it to be held stationary. Its weight acts downward, and the electric force acts upwards to oppose the weight.
Original post by 1234kelly
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 ?
(edited 1 year ago)