What is the charge that can be stored on this capacitor

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The discussion revolves around calculating the charge stored in a parallel-plate capacitor with specific dimensions and a dielectric material. The initial calculation yielded a capacitance of 200 pF, but subsequent corrections indicated a recalculated capacitance of approximately 54.18 pF. Using this corrected value, the charge stored when connected to a 1.5-V battery was recalculated to be around 81.27 pC. Participants confirmed that the final charge value appeared accurate based on the revised capacitance. The focus remained on ensuring the calculations adhered to the correct formulas and values.
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Homework Statement



A parallel-plate capacitor has dimensions 3.5cm x 5.2 cm. The plates are separated by a 1.1-mm thickness of paper (dielectric constant k = 3.7). What is the charge that can be stored on this capacitor, when connected to a 1.5-V battery? (E0 = 8.85 x 10^-12 C^2/Nm^2)

Homework Equations



C=Keo(A/d)
Qmax=(Cdelta)Vmax

The Attempt at a Solution

C=Keo(A/d)
=3.7(8.85E-12)(18.2E-4/1.1E-3)
=200E-12 F

Qmax=C(deltaVmax)
=(200E-12)(1.5V)
=300E-12 C

300E-12 C does not seem like the right answer
 
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Hi ziplock2k,

ziplock2k said:

Homework Statement



A parallel-plate capacitor has dimensions 3.5cm x 5.2 cm. The plates are separated by a 1.1-mm thickness of paper (dielectric constant k = 3.7). What is the charge that can be stored on this capacitor, when connected to a 1.5-V battery? (E0 = 8.85 x 10^-12 C^2/Nm^2)

Homework Equations



C=Keo(A/d)
Qmax=(Cdelta)Vmax

The Attempt at a Solution




C=Keo(A/d)
=3.7(8.85E-12)(18.2E-4/1.1E-3)
=200E-12 F

I don't think those numbers give 200e-12 F.
 
C=Keo(A/d)
=3.7(8.85E-12)(18.2E-4/1.1E-3)
=54.18E-12

Qmax=C(deltaVmax)
=(54.18E-12)(1.5V)
=300E-12 C

81.27E-12 C

Is my new calculations right?
 
ziplock2k said:
C=Keo(A/d)
=3.7(8.85E-12)(18.2E-4/1.1E-3)
=54.18E-12

Qmax=C(deltaVmax)
=(54.18E-12)(1.5V)
=300E-12 C

81.27E-12 C

Is my new calculations right?

That last number looks right to me.
 
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