Calculating Charge Stored on Parallel Plates with Non-Conducting Water

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The discussion focuses on calculating the charge stored on parallel plates with non-conducting water between them. The user initially attempts to use the formula Q = CV but makes a mistake in converting the area from cm² to m². The correct conversion for the area should be noted as 4.67 cm² equals 4.67 x 10^-4 m². After correcting the area conversion, the user recalculates the charge, leading to a more accurate result. Proper unit conversion is crucial for accurate calculations in physics problems.
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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.
 
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Tekmachine said:
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! :smile:

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

erm … 4.67 cm2 isn't (4.67 * 10^-2) m2. :smile:
 
i feel so dumb, lol thanks!
 
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