Calculating Charge Stored on Parallel Plates with Non-Conducting Water

[Tekmachine]

Homework Statement

Two parallel plates, each of area 4.67 cm2, are separated by 3.93 mm
with purified non-conducting water between them. A voltage of 5.39 V
is applied between the plates.

Find the charge stored on each plate. Answer in units of nC.

Homework Equations

$$C = K\frac{(\varepsilon_{0}A)} {d}}$$

$$Q = CV$$

The Attempt at a Solution

Q = [ 5.39 * 80 * (8.85 * 10^-12) (4.67 * 10^-2) ] / (3.93 * 10^-3)

Q = 4.534 * 10^-8

When I convert that to nC I get 45.34 nC. What went wrong? =(

Also, sorry for the horrible formatting! I'm having database errors when trying to use too much LaTex.

 

[tiny-tim]

Two parallel plates, each of area 4.67 cm² …

Q = [ 5.39 * 80 * (8.85 * 10^-12) (4.67 * 10^-2) ] / (3.93 * 10^-3)

Hi Tekmachine!

(no need to use LaTeX … just use the X² and X₂ tags above the Reply field. )

erm … 4.67 cm² isn't (4.67 * 10^-2) m².

 

[Tekmachine]

i feel so dumb, lol thanks!