The 2.00 cm^3 volume between the plates of a 249 pF parallel plate capacitor is
ID: 1530046 • Letter: T
Question
The 2.00 cm^3 volume between the plates of a 249 pF parallel plate capacitor is filled with a dielectric. The area of each plate is 0.00500 m^2 and the plates are 400 mu m apart. Find the permittivity of the dielectric. To one significant figure, find the maximum energy density of an electric field in production-quality polystyrene (Table 24.2. page 802). A 3.0 kF capacitor stores 3.0 watt-hours = 10.8 kJ in its electric field. What is its stored charge? Find the distance (in mu m) between the plates of a parallel plate capacitor if, in the dielectric completely filling that region, an electric field of magnitude 88.8 MV/m gives a potential difference of 22.2 kV between the plates. In vacuum, the potential difference between the non-parallel plates of a capacitor is given by (0.0345 m^-1)Q/epsilon_0. Calculate the numerical value of the capacitance of this capacitor after neoprene at 20 degree C completely fills the volume between the plates so there is no change in symmetry from the vacuum case. (Use Table 24.1. page 798.) Step 1: With the dielectric in place, we simply replace epsilon_0 in the vacuum expression for the potential difference with_ epsilon_0. Substituting neoprene's K from Table 24.1 then gives potential difference of (0.0345 m^-1)Q/()(F/m). (your Step 2 denominator must contain the numerical values of K and epsilon_0). Step 3: Now show all of your numbers and the necessary algebraic cancellation as you substitute your Step 2 expression into your defining equation, giving the capacitance =Explanation / Answer
1) Here, 249 * 10-12 = (K * 8.854 * 10-12 * 0.005)/(400 * 10-6)
=> permittivity of dielectric , K = 2.25
3) Here, 10.8 * 1000 = 1/2 * Q2/(3000)
=> Q = 8050 C
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