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)

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cjmejiab cjmejiab · Answer 1 · 2019-12-26T05:59:18+01:00

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

[tex]Q = CV[/tex]

Where,

C = Capacitance

V = Voltage,

Replacing we have

[tex]Q = 0.02*12[/tex]

[tex]Q = 0.24C[/tex]

When dialectics is increased the new capacitance is

[tex]C' = KC[/tex]

C' = 3.2* 0.02

[tex]C' = 0.064F[/tex]

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

[tex]E = \frac{Q^2}{2C'}[/tex]

[tex]E = \frac{0.24^2}{2*0.064}[/tex]

[tex]E = 0.45J[/tex]

Therefore the Energy stored is 0.45J