A parallel plate capacitor has charge $5 \times 10^{-6} \mathrm{C}$. A dielectric slab is inserted between…

A parallel plate capacitor has charge $5 \times 10^{-6} \mathrm{C}$. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{-6} \mathrm{C}$ then the dielectric constant of the slab is ________.

Solution

$\begin{aligned} & \mathrm{Q}_{\mathrm{in}}=\mathrm{Q}\left(1-\frac{1}{\mathrm{~K}}\right) \\ & 4 \times 10^{-6}=5 \times 10^{-6}\left(1-\frac{1}{\mathrm{~K}}\right) \\ & 1-\frac{1}{\mathrm{~K}}=\frac{4}{5} \\ & \mathrm{~K}=5\end{aligned}$

Asked in: JEE Main 2025 (07 Apr Shift 2)

Practice more Electrostatics questions on Aicharya