Question

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

Answer 1

Answer:

refer to the attachment....

Answer 2

Answer:

Capacitance = $5C$

Energy stored = $\dfrac 15 W$.

Explanation:

The capacity or the capacitance of a capacitor is given by

$C=\dfrac{A\epsilon}{d}$.

The energy stored in the capacitance is given by

$W=\dfrac{Q^2}{2C}$.

where,

• $\epsilon$ = electric permittivity of the medium that fills the region between the plates of the capacitor.

• $A$ = area of the plates of the capacitor.

• $d$ = distance between the plates of the capacitor.

• $V$ = voltage across the plates of the capacitor.

When the air fills the gap between the plates, then

$C=\dfrac{A\epsilon_o}{d}$

When the air is replaced by the glass which have dielectric constant $k=5$, then the new capacity of the capacitor is given by

$C'=\dfrac{A\epsilon}{d}\\ = \dfrac{A(k\epsilon_o)}{d}\ \ \ \ \because \epsilon = k\epsilon_o\\ =k\dfrac{A\epsilon_o}{d}\\ = kC = 5C.$

Since, the charge remains conserved, therefore,  the new energy stored is given by

$W'=\dfrac {Q^2}{2C'} \\= \dfrac {Q^2}{2(5C)} \\= \dfrac 15 \dfrac {Q^2}{2C}  \\= \dfrac 15W$