How Do You Solve This Capacitor Problem with Aluminum Electrodes?

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The problem involves calculating the capacitance of two aluminum electrodes with a diameter of 5.0 cm and a separation of 0.50 mm, connected to a 200 V battery. The capacitance formula C = εA/d was applied, with the area calculated as π * (d^2/4). The user obtained a capacitance of 139 pF, but the online homework platform indicated this result was incorrect. The discussion highlights a potential error in the area calculation, suggesting a review of the formula used for the area of the electrodes. Accurate calculations are crucial for determining the correct capacitance value.
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Homework Statement


Two 5.0-cm-diameter aluminum electrodes are spaced 0.50 mm apart. The electrodes are connected to a 200 V battery.

Homework Equations


C = εA/d

The Attempt at a Solution


So for my solution I took the area of the plate, ((.052π) * 8.85*10-12) / .5*10-3 and I got an answer of 1.39 * 10-10 F or 139 pF. The website the homework is done through is saying this is wrong. not too sure where the mistake is...
 
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Area of electrode = pi * d^2 /4

Looks like you used pi * d^2
 

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