Parallel plate capacitor and charge density

[tony873004]

Homework Statement

A parallel plate capacitor with A=125 cm² and d=1.25 mm is charged to a potential difference of 10.0 V. What is the charge density on each plate?

Homework Equations

$$V = \frac{{\sigma d}}{{\varepsilon _0 }}\,\,\,\, \Rightarrow \,\,\,\,V\varepsilon _0 = \sigma d\,\,\,\, \Rightarrow \,\,\,\,\sigma = \frac{{V\varepsilon _0 }}{d}$$

The Attempt at a Solution

Plugging in numbers, I get 8.5*10^-8C/m². But I never used the Area. My units seem to work. Did they throw in irrelavant data? Shouldn't I only need Area if I want total charge?

Should my answer be half of what I computed, because it is divided between 2 plates?

 

[Hootenanny]

As far as I'm concerned your solution is correct. As you correctly point out, one only needs the area to calculate the total charge on each plate. Further, you do not need to halve your answer since the surface charge density is calculated for each plate. You can check your solution by placing a Gaussian surface around one plate. Perhaps http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c2" would be helpful.

 

[tony873004]

Thanks for the feedback and the link!