Question

The capacitor is then disconnected from the 12V battery and a dielectric with a dielectric constant of k is inserted between the plates. How much energy will be stored in the capacitor after inserting the dielectric? Assign values for C (0.02 F) and k (3.2)

Answer 1

To solve this problem we will apply the concepts related to the energy stored in a capacitor

and capacitance from the dialectic. Finally we will determine the energy stored according to the load and capacitance.

The charge store in capacitor is

$Q = CV$

Where,

C = Capacitance

V = Voltage,

Replacing we have

$Q = 0.02*12$

$Q = 0.24C$

When dialectics is increased the new capacitance is

$C' = KC$

C' = 3.2* 0.02

$C' = 0.064F$

Finally with this values we can calculate the energy stored, which is given as,

$E = \frac{Q^2}{2C'}$

$E = \frac{0.24^2}{2*0.064}$

$E = 0.45J$

Therefore the Energy stored is 0.45J