- #1
123ryoma12
- 8
- 1
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.