Question

A parallel-plate air capacitor has a capacity '\(C\)', a separation '\(x\)', a dielectric material with a dielectric constant '\(k\)' between plates, and a potential difference of '\(V\)' volt applied between plates. The force of attraction between the plates is

• A. \(\frac{kCV^{2}}{2x}\)
• B. \(\frac{kCV^{2}}{2}\)
• C. \(\frac{kCV}{2x}\)
• D. \(\frac{kCV^{2}}{x}\)

Hint:

Find the charge on a capacitor filled with a dielectric and the field acting on one plate.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
With the dielectric filling the gap, the capacitance becomes \(kC\). The charge on the plates is \(Q=kCV\).

Step 2: Field
The potential difference is \(V\) across the gap \(x\), so the field is \(E=\dfrac{V}{x}\).

Step 3: Force
The pull on one plate is \(F=\dfrac{1}{2}QE=\dfrac{1}{2}kCV\cdot\dfrac{V}{x}=\dfrac{kCV^{2}}{2x}\).

Final Answer:
The force of attraction is kCV squared over 2x. \[ \boxed{\text{(A) }\dfrac{kCV^{2}}{2x}} \]