When a parallel plate capacitor is charged up to 95 V , its capacitance is C. If a dielectric slab of…

When a parallel plate capacitor is charged up to 95 V , its capacitance is C. If a dielectric slab of thickness 2 mm is inserted between plates and distance between the plates is increased by 1.6 mm such that the same potential difference is maintained. The dielectric constant of the material (slab) is
  1. 2.4
  2. 4.5
  3. 5.0
  4. 9.0

Solution

When the battery is removed,
$\begin{aligned} & Q=\text { constant } \\ & \Rightarrow C_1 V=C_2 V \Rightarrow C_1=C_2 \\ & \Rightarrow \frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{(d+1.6)-t\left(1-\frac{1}{k}\right)} \\ & \therefore d=(d+1.6)-2\left(1-\frac{1}{k}\right) \\ & \therefore \quad K=5 \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

Practice more Electrostatics questions on Aicharya