Question

The parallel plate condenser has plates of area 200 cm sq. And separation of 0.05 cm .the space between plates have been filled with a dielectric having k=8 and then charged to 300 volts.The stored energy is?

Answer 1

this may help u........

Answer 2

Answer:

$1.274\times 10^{-4}\ J$

Explanation:

Given:

• $\epsilon_o$ = permittivity of free space = $8.85\times 10^{-12}\ F/m$
• $k$ = dielectric constant of the material filled = 8
• A = area of each plate of the capacitor = $200\ cm^2=2\times 10^{-2}\ m^2$
• d = distance between the plates of the capacitors = $0.05\ cm = 5\times 10^{-4}\ m$
• V = voltage supplied between the capacitors = 300 V

Assume:

• C = capacitance of the capacitor
• U = energy stored in the capacitor

Capacitor is a device which is used in an electric circuit in order to store the energy in the form of charge.

The capacitance of a capacitor is given by:

$C = \dfrac{k\epsilon_o A}{d}\\\Rightarrow  C = \dfrac{8\times 8.85\times 10^{-12}\times 2\times 10^{-2} }{5\times 10^{-4}}\\\Rightarrow  C =2.832\times 10^{-9}\ F$

Now energy stored in the capacitor is:

$U = \dfrac{1}{2}CV^2\\\Rightarrow U = \dfrac{1}{2}\times 2.832\times 10^{-9}\times 300^2\\\Rightarrow U = 1.274\times 10^{-4}\ J$

Hence, the energy stored in the capacitor is $1.274\times 10^{-4}\ J$.