Question

An air-gap parallel plate capacitor of capacitance c0 = 20 nf is connected to a battery with voltage v = 12 v. while the capacitor remains connected to the battery, we insert a dielectric (κ = 2.6) into the gap of the capacitor, filling one half of the volume as shown below. what is u, the energy stored in the this capacitor?

Answer 1

3.7 * 10
`^(-5) J`

Step-by-step explanation:

Thinking process:

Let the energy be calculated by the following:

`U = (1)/(2)CV^(2)`

where V is the voltage applied across the load.

C is the capacitance

In case of a dielectric, the capacitance is given by the following equation:

`C = kC_(0)`

where
`C_(0)` is the capacitance in vacuum. So, the energy stored becomes:

`U = (1)/(2) (kC_(o))V^(2)`

Then, k = 2.6 ,
`C_(0) = 20 nF`, and V = 12 V

Therefore, in the problem, the energy stored becomes:

`U = (1)/(2) (2.6 * 20*10^(-9)) (12)^(2) \\ = 3.7 * 10^(-5) J`