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2. Electric Potential and Capacitance
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A parallel plate capacitor with air between the plate has a capacitance of $15 pF$. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant $3.5.$ Then the capacitance becomes $\frac{ x }{4}\,pF$.The value of $x$ is $............$
A
$10.5$
B
$1.05$
C
$105$
D
$108$
(JEE MAIN-2023)
Solution
$C _0=\frac{\in_0 A }{ d }=15\,pF$
$C =\frac{ K \in_0 A }{2 d }=\frac{3.5}{2} \times 15\,pF$
$=\frac{105}{4}\,pF$
Standard 12
Physics
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