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2. Electric Potential and Capacitance
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The capacity of a parallel plate condenser is $15\,\mu \,F$, when the distance between its plates is $6 \,cm$. If the distance between the plates is reduced to $2\, cm$, then the capacity of this parallel plate condenser will be......$\mu \,F$
A
$15$
B
$30$
C
$45$
D
$60$
Solution
(c) $C \propto \frac{1}{d} \Rightarrow $$\frac{{{C_1}}}{{{C_2}}} = \frac{{{d_2}}}{{{d_1}}}$ $⇒$ $\frac{{15}}{{{C_2}}} = \frac{2}{6} ⇒ C_2 = 45\,\mu F$
Standard 12
Physics
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