The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $C_{0}$ and $W_{0}$ . If the air between the plates is replaced by glass (dielectric constant $= 5$ ) find the capacitance of the consider and the energy stored in it.

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Solution

The capacity in a parallel plate condenser with air between its plates is $C_{0} = \frac{k ε_{0} A}{d}$

If the air between the plates is replaced by glass than

$C = \frac{5 k ε_{0}}{d}$

$C = 5 C_{0}$

Energy stored in a parallel plate condenser with air between its plates is $W_{0}$

$W_{0} = \frac{1}{2} C V^{2}$
When glass is inserted between the plates then energy stored is

$E = \frac{1}{2} \times 5 C_{0} \times V^{2}$

$E = 5 W_{0}$

Both the values of capacitance and energy stored after inserting glass will increase.

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The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively C0 and W0. If the air between the plates is replaced by glass (dielectric constant=5) find the capacitance of the consider and the energy stored in it.

The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $C_{0}$ and $W_{0}$ . If the air between the plates is replaced by glass (dielectric constant $= 5$ ) find the capacitance of the consider and the energy stored in it.