JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 25)
A parallel plate capacitor with air between the plate has a capacitance of 15pF. 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 ____________.
Answer
105
Explanation
Initially
$$ \frac{\varepsilon_{0} A}{d}=15 \times 10^{-12} \mathrm{~F} $$
Finally
$$ \begin{aligned} & \frac{3.5 \varepsilon_{0} A}{2 d}=\frac{x}{4} \times 10^{-12} \mathrm{~F} \\\\ & \therefore \frac{3.5}{2} \times 15=\frac{x}{4} \\\\ & \Rightarrow x=\frac{3.5 \times 15 \times 4}{2}=105 \end{aligned} $$Comments (0)
