Electric Field Strength: 2009 HSC Exam Question Explained | Australia

[123ryoma12]

This is from the 2009 HSC exam. (I'm in Australia)
I checked the answers and found that electric field strength was
E = 100/0.10 = 1000
My question is, what would the electric field strength be if the positively charged plate was +100V instead of 0V
Would it be 200/0.1 = 2000?
Is the formula E = The difference in volts between the two plates / distance.
Please help this has been bugging me.

 

[Orodruin]

E = 100/0.10 = 1000

No it was not. The electric field strength is a dimensionful quantity and you simply cannot quote it as just a number without a unit. The field strength is 1000 V/m = 1 kV/m.

Would it be 200/0.1 = 2000?

It would be 2000 V/m.

 

[123ryoma12]

I just realized that I shouldn't have posted here. Sorry about that.
But another question
http://www.regentsprep.org/Regents/physics/phys03/aparplate/plate3.gif
In this gif where
E = V/d
Shouldn't it be E = 2V/d
as there is voltage going to the positive and negatively charged plate for example
if the battery has 10V
The negatively charged would be -10V and the positively charge would be 10V

 

[davenn]

Shouldn't it be E = 2V/d
as there is voltage going to the positive and negatively charged plate for example
if the battery has 10V
The negatively charged would be -10V and the positively charge would be 10V

why would you think that ?
There isn't a 20V difference across the plates
one terminal of the battery, the positive, is +10V relative to the 0V of the negative terminalDave

 

[123ryoma12]

Oh ok thanks. I just thought that for some reason. I didn't really know how the battery worked.