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

The energy stored in a capacitor is:
$U= \frac{1}{2} C V^2$
where V is the voltage applied, and C is the capacitance. In case of a dielectric, the capacitance is given by
$C=k C_0$
where $C_0$ is the capacitance in vacuum. So, the energy stored becomes 
$U= \frac{1}{2} (k C_0) V^2$

In our problem we have k=2.6, $C_0 = 20 nF=20 \cdot 10^{-9} F$ and V=12 V. therefore the energy stored is
$U= \frac{1}{2} (2.6\cdot 20\cdot 10^{-9}F)(12 V)^2=3.7 \cdot 10^{-5}J$

Answer 2

Answer:

3.7 * 10$^{-5} J$

Explanation:

Thinking process:

Let the energy be calculated by the following:

$U = \frac{1}{2}CV^{2}$

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:

$U = \frac{1}{2} (kC_{o})V^{2}$

Then, k = 2.6 , $C_{0} = 20 nF$, and V = 12 V

Therefore, in the problem, the energy stored becomes:

$U = \frac{1}{2} (2.6 * 20*10^{-9}) (12)^{2} \\ = 3.7 * 10^{-5} J$