Energy of a Capacitor in the Presence of a Dielectric

[vachan]

A dielectric-filled parallel-plate capacitor has plate area of 20.0 and plate separation of 5.00 . The capacitor is connected to a battery that creates a constant voltage of 7.50 . The dielectric constant is 2.00. Throughout the problem, use 8.85×10−12 .

The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy of the capacitor at the moment when the capacitor is half-filled with the dielectric.Express your answer numerically in joules.

I got confused ... and i am lost.. what equation should i use?

 

[rl.bhat]

plate area of 20.0 and plate separation of 5.00 .
What are the units of these quantities?

 

[Moetasim]

I can provide an explanation and guide you through the process of finding the energy of the capacitor in this scenario.

Firstly, it is important to understand that the energy of a capacitor is given by the equation: E = 1/2 CV^2, where C is the capacitance and V is the voltage.

In this problem, we are given the plate area (A) and plate separation (d) of the capacitor, which can be used to calculate the capacitance using the formula: C = ε0 A/d, where ε0 is the permittivity of free space (8.85×10−12).

Plugging in the values, we get: C = (8.85×10−12)(20.0)/5.00 = 3.54×10−11 F.

Next, we are provided with the voltage (V) of 7.50 V and the dielectric constant (κ) of 2.00. When the capacitor is half-filled with the dielectric, the effective capacitance (C') is given by: C' = κC = (2.00)(3.54×10−11) = 7.08×10−11 F.

Now, we can use the formula for energy of a capacitor to calculate the energy at this moment: E = 1/2 C'V^2 = 1/2 (7.08×10−11)(7.50)^2 = 1.99×10−9 J.

Therefore, the energy of the capacitor at the moment when it is half-filled with the dielectric is 1.99×10−9 J.

I hope this explanation helps you understand the problem better and guides you through the steps to find the solution.