NEET Physics Question - Capacitors - Capacitor With Dielectrics

An air-filled parallel-plate capacitor has a capacitance of \(1\text{ pF}\). The plate separation is then doubled and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes \(2\text{ pF}\). The dielectric of the wax is

0.25

0.5

2.0

4.0

Solution:

Initial capacitance \(C_0 = \frac{\varepsilon_0 A}{d} = 1\text{ pF}\). Doubling the distance and inserting a dielectric \(k\) makes the capacitance \(C = \frac{k \varepsilon_0 A}{2d} = \frac{k}{2} C_0\). Since \(C = 2\text{ pF}\), we obtain \(2 = \frac{k}{2}(1) ⇒ k = 4\).

Rankers Physics

Dielectric Constant of Wax

Solution:

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