7) Two circular metal plates, each of radius \(r\), are kept parallel to each other at a distance \(d\) apart, forming a capacitor with capacitance \(C_1\). If the radius of each plate is increased to \(\sqrt{2}\) times the original radius and their separation is decreased to half the initial value, the new capacitance is \(C_2\). The ratio of the capacitances \(C_1:C_2\) is:
Solution:
For a parallel plate capacitor, the capacitance is given by: \[ C = \frac{\epsilon_0 A}{d}, \] where \(A\) is the area of the plate. For circular plates, \(A = \pi r^2\).
The initial capacitance is: \[ C_1 = \frac{\epsilon_0 \pi r^2}{d}. \] When the radius is increased to \(\sqrt{2}r\), the new area becomes: \[ A' = \pi (\sqrt{2}r)^2 = 2\pi r^2, \] and when the separation is decreased to \(\frac{d}{2}\), the new capacitance is: \[ C_2 = \frac{\epsilon_0 (2\pi r^2)}{d/2} = \frac{4\epsilon_0 \pi r^2}{d}. \]
Thus, the ratio of the capacitances is: \[ \frac{C_1}{C_2} = \frac{\epsilon_0 \pi r^2/d}{4\epsilon_0 \pi r^2/d} = \frac{1}{4}. \] So, the ratio \(C_1:C_2\) is \(1:4\), making option C the correct answer.
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