If K is dielectric constant of a medium and $\in_{0}$ permittivity of free space, then the energy stored per unit volume of the medium is given by

\frac{1}{2} K_{2} \frac{\in_{0}^{2}}{E}

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\frac{1}{2} K \in_{0}^{2} E

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\frac{1}{2} K \in_{0} E^{2}

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\frac{1}{2} K \in_{0} E

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Solution

The correct option is B $\frac{1}{2} K \in_{0} E^{2}$
The permittivity of dielectric medium is $ϵ = K ϵ_{0}$
The energy stored per unit volume in the dielectric medium $=$ energy density $= \frac{1}{2} ϵ E^{2} = \frac{1}{2} K ϵ_{0} E^{2}$ where $E$ is the electric filed in that medium.

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K

\to E

ϵ_{0}

E

K (i n J m^{- 3})

$X = x$	0	1	2	3	4	5	6	7
$P (X = x)$	$0$	$K$	$2 K$	$2 k$	$3 K$	$K^{2}$	$2 K^{2}$	$7 K^{2} + k$

P (0 < X < 5) =

f (x) = \frac{a_{2 k} x^{2 k} + a_{2 k - 1} x^{2 k - 1} + . . . . + a_{1} x + a_{0}}{b_{2 k} x^{2 k} + b_{2 k - 1} x^{2 k - 1} + . . . . + b_{1} x + b_{0}}

k

a_{i}, b_{i} \in R

a_{2 k} \neq 0, b_{2 k} \neq 0

b_{2 k} x^{2 k} + b_{2 k - 1} x^{2 k - 1} + . . . . . + b_{1} x + b_{0} = 0

$X = x$	$0$	1	2	3	4	5	6	7
$P (X = x)$	$0$	$k$	$2 k$	$2 k$	$3 k$	$K^{2}$	$2 k^{2}$	$7 k^{2} + k$

P (0 < X < 5) =

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