Sunday
May 19, 2013

# Homework Help: Physics

Posted by Robert Soos on Sunday, February 19, 2012 at 11:13pm.

When an air capacitor with a capacitance of 380 nF (1 {\rm nF} = 10^{-9}\;{\rm F}) is connected to a power supply, the energy stored in the capacitor is 1.95×10−5 J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.62×10−5 J.
A) What is the potential difference between the capacitor plates?
B)What is the dielectric constant of the slab?

• Physics - drwls, Sunday, February 19, 2012 at 11:31pm

Stored Energy = (1/2)CV^2
It increases by a factor (2.62+1.95)/1.95 = 2.344
A) Since V is constant, C increases by the same factor, to
2.344*380 nF = 891 nF
B) The dielectric constant is the factor that increased the capacitance, 2.344

• Physics - Vivek, Monday, February 20, 2012 at 5:30am

Wgat is energy

• Physics (to Vivek) - drwls, Monday, February 20, 2012 at 5:33am

Nonsense.

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