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 = (Q^2)/(2C')`

`E = (0.24^2)/(2*0.064)`

`E = 0.45J`

Therefore the Energy stored is 0.45J