Question

The plates of parallel plate capacitor are charged upto 100V. A 2 mm thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. The dielectric constant of the plate is

• A. 5
• B. 1.25
• C. 4
• D. 2.5

Answer 1

Answer: (A)

In air the potential difference between the plates $V_{air}=\frac{\sigma }{{\epsilon }_0}\cdot d$ ... (i) In the presence of partially filled medium potential difference between the plates $V_m=\frac{\sigma }{{\epsilon }_0}\left(d-t+\frac{t}{K}\right)$ ? (ii) Potential difference between the plates with dielectric medium and increased distance is $V_m=\frac{\sigma }{{\epsilon }_0}\left\{(d+d)-t+\frac{t}{k}\right\}$ ... (iii) According to question $V_{air}=V_m$ which gives $K=\frac{t}{t-d}$ Hence, $K=\frac{2}{2-1.6}=5$