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A parallel plate capacitor with air between the plate has a capacitance of \mathrm{15pF}. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant \mathrm{3.5}.Then the capacitance becomes \mathrm{\frac{x}{4}pF}. The value of \mathrm{x} is _________

Option: 1

105


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \mathrm{C}=\frac{\mathrm{A} \varepsilon_0}{\mathrm{~d}} \\ \end{aligned}

\begin{aligned} & \mathrm{C}=\frac{\mathrm{KA \varepsilon _{0 }}}{2 \mathrm{~d}} \\ \end{aligned}

\begin{aligned} & =\frac{\mathrm{KC}}{2} \\ \end{aligned}

\begin{aligned} & =\frac{3.5 \times 15}{2} \\ \end{aligned}

\begin{aligned} & =\frac{105}{4} \\ & =105 \end{aligned}

Posted by

Devendra Khairwa

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