How much charge stored in capacitor ?

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The discussion focuses on calculating the maximum voltage and charge stored in a capacitor with a paper dielectric. Given the area of the plates (100 cm²), thickness (0.005 cm), and dielectric constant (2.5), the capacitance can be determined using the formula C = ε*A/d, where ε is the permittivity. The maximum voltage is constrained by the electric field strength of 5 x 10^5 V/cm. Once the capacitance is calculated, the charge stored can be found using Q = V*C. The calculations provide insights into capacitor behavior with dielectric materials.
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A paper is placed between two plates of copper , area 100 cm^2. The thickness of the paper 0.005cm and its dielectric constant is 2.5 . If the paper can bear an electric field of 5 x 10^5 V/cm then calculate the maximum voltage upto which the capacitor can be charged.How much charge will be stored on the capacitor ?



My attempt -> A=100cm^2
K=2.5
5x10^5=9x10^9 x q / r
 
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ixtra said:
A paper is placed between two plates of copper , area 100 cm^2. The thickness of the paper 0.005cm and its dielectric constant is 2.5 . If the paper can bear an electric field of 5 x 10^5 V/cm then calculate the maximum voltage upto which the capacitor can be charged.How much charge will be stored on the capacitor ?

My attempt -> A=100cm^2
K=2.5
5x10^5=9x10^9 x q / r

Welcome to PF.

First figure the capacitance
ε = 2.5*εo
C = ε*A/d = 2.5*8.8*10-12*(.01 m2)/.00005m

Then you can use that result to figure the charge at max V with

Q = v*C
 
thanks
 
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