Charge density - parallel plates

[pat666]

Homework Statement

5. Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm.
a) If the surface charge density for each plate has a magnitude of 47.0 nCm^-2, what is the magnitude of the electric field intensity in the region between the plates?
b) What is the potential difference between the two plates? (2 marks)

Homework Equations

The Attempt at a Solution

The wording here confuses me a bit, is Electric field intensity something different to the electric field?
I've done E=4.7*10^-9/8.85E-12 =530.8N/C but is this the field intensity? Part b I am not sure about, I need some help for it to.

Thanks

 

[Doc Al]

Electric field and electric field intensity: same thing. (Check your arithmetic: is the charge density 4.7 or 47?)

For part b: You'll need to use the distance.

 

[pat666]

47, so E=5310.7N/C
Then V=kq/r but I don't know q so how would I do that?

 

[Doc Al]

Then V=kq/r but I don't know q so how would I do that?

That formula is for a point charge, which is not relevant here. What's the general relationship between field and potential?

 

[pat666]

found it, V=Ed

so V=116.8V?

 

[Doc Al]

You got it.

 

[pat666]

Thanks