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roxiney
Joined: 15 Sep 2010 Posts: 5 Location: north carolina

Posted: Wed Sep 22, 2010 11:24 pm Post subject: Work and Potential difference 


A 3.5nF parallel plate capacitor with a sheet of Mylar (κ = 3.1) filling the space between the plates is charged to a potential difference of 228.0 V and is then disconnected. How much work is required to completely remove the sheet of Mylar from the space between the two plates? 

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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Thu Sep 23, 2010 8:12 pm Post subject: 


For this question you need to consider the energy difference between when the capacitor is initially charged and when the dielectric is pulled out.
Initially, the capacitor is charged to its voltage and then the battery is disconnected. When the battery is disconnect the charge still remains on the capacitor. The energy stored in the capacitor is given by E= 1/2 CV^2. Calculate this.
Another formula for the capacitance of a parallel plate capacitor is:
C= (epsilon_0 * epsilon_r * A)/d where epsilon_0 is the permittivity of free space, epsilon_r is the relative permittivity of the dielectric, A is the area of the plates and d is their separation.
Since the only thing that changes when the dielectric is pulled out is the relative permittivity (from 3.1 to 1) it means that the capacitance must decrease by a factor of (1/3.1). So the new value of C is now 3.5nF/3.1
Calculate the energy with this new capacitance. Remember to use the correct units.
The work required, assuming no extraneous energy losses, will be the difference between the two energies. 

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