Electric Field and Force between two plates [general question]

[Samurai44]

Greetings, I am just confused in something , Is there a difference when calculating the electric field , or a force exerted on a charge between two plates , if the two plates were :
a) both plates are charged ( connected to a power source) ,
OR
b) both plates are charged ( connected to a power source), but the negative plate is earthed , i.e 0 V at the negative plate .

 

[sophiecentaur]

It only depends upon the Potential Difference (as long as there is no other structure nearby). SO connecting between +6V and -6V will give the same result as +12V and earth.

 

[Samurai44]

It only depends upon the Potential Difference (as long as there is no other structure nearby). SO connecting between +6V and -6V will give the same result as +12V and earth.

i mean let's suppose the voltage is 6V at the + plate , but the negative plate is 0v because its earthed

 

[Doc Al]

i mean let's suppose the voltage is 6V at the + plate , but the negative plate is 0v because its earthed

In that case the potential difference will be 6 volts.

 

[Samurai44]

In that case the potential difference will be 6 volts.

so when finding a force on a charge(e .g an electron) between the plates , for example at the mid , using this formula "F=e V/d" ,, do we use the value of V as 6 ?

 

[sophiecentaur]

so when finding a force on a charge(e .g an electron) between the plates , for example at the mid , using this formula "F=e V/d" ,, do we use the value of V as 6 ?

Field is Volts per Metre. If you have 6V PD between the plates then it's 6V for the formula.
I am just wondering if you could be worrying about the distinction between Potential and Potential Difference. When dealing with situations on Earth then you take Earth Potential as Zero and 'Potential' is relative to Earth. If the minus terminal of the supply is connected to Earth , the 'Potential' of the + terminal is 6V - and vice versa, when connecting the other way round.