Question

The plates of a parallel plate capacitor are charged up to $100$ volt. A $2mm$ thick plate is inserted between the plates, then to maintain
the same potential difference, the distance between the capacitor plates is increased by $1.6mm$ . The dielectric constant of the plate is :

• A. $5$
• B. $1.25$
• C. $4$
• D. $2.5$

Answer 1

Answer: (A)

Since, the potential difference between the plates of capacitor remains same and charge on the capacitor is also same. Therefore, the capacitance of parallel plate capacitor should remain same after insertion of dielectric slab and increasing the separation between the plates.
Initial capacitance of capacitor, $C=\frac{\epsilon _{0} A}{d}$ .
Final capacitance of capacitor, $C=\frac{\left(\epsilon \right)_{0} A}{D - t \left(1 - \frac{1}{k}\right)}=\frac{\left(\epsilon \right)_{0} A}{d + 1 . 6 - 2 \left(1 - \frac{1}{k}\right)}$
From above equations,
$\frac{\left(\epsilon \right)_{0} A}{d + 1 . 6 - 2 \left(1 - \frac{1}{k}\right)}=\frac{\left(\epsilon \right)_{0} A}{d}\Rightarrow d=d+1.6-2+\frac{2}{k}\Rightarrow k=5$