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2. Electric Potential and Capacitance
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The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively ${C_o}$ and ${W_o}$. If the air is replaced by glass (dielectric constant $= 5$ ) between the plates, the capacity of the plates and the energy stored in it will respectively be
A
$5{C_o},\;5{W_o}$
B
$5{C_o},\;\frac{{{W_0}}}{5}$
C
$\frac{{{C_o}}}{5},\;5{W_o}$
D
$\frac{{{C_o}}}{5},\frac{{{W_o}}}{5}$
Solution
(b) When a dielectric $K$ is introduced in a parallel plate condenser its capacity becomes $K$ times. Hence $C' = 5{C_0}$. Energy stored ${W_0} = \frac{{{q^2}}}{{2{C_0}}}$
$W' = \frac{{{q^2}}}{{2C'}} = \frac{{{q^2}}}{{2 \times 5{C_0}}}$ $==>$ $W' = \frac{{{W_0}}}{5}$
Standard 12
Physics
