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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 _____.
Correct Answer: 105
$$C_1 = \frac{\varepsilon_0 A}{d} = 15 \text{ pF}$$
$$C_2 = \frac{K_2\varepsilon_0 A}{d_2} = \frac{3.5 \cdot \varepsilon_0 A}{2d}$$
$$C_2 = \frac{3.5}{2} \times 15 = \frac{52.5}{2} = 26.25 \text{ pF}$$
$$x = 26.25 \times 4 = 105$$
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