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Arada [10]
3 years ago
6

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)
Physics
1 answer:
sergij07 [2.7K]3 years ago
7 0

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

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