Question

The equivalent capacitance between 'P' and 'Q' is [A-area of each plate, d-separation between the plates, \(\epsilon _{0}\) -permittivity of free space]

• A. \(\frac{A\epsilon _{0}}{d}\)
• B. \(\frac{2A\epsilon _{0}}{d}\)
• C. \(\frac{3A\epsilon _{0}}{d}\)
• D. \(\frac{4A\epsilon _{0}}{d}\)

Hint:

The three slabs are side by side between the same plates, so they act as parallel capacitors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Each slab fills one third of the plate area (A/3) with the same plate separation \(d\), so the three capacitors are in parallel. The dielectric constants are \(K_1=1\), \(K_2=2\), \(K_3=3\).

Step 2: Individual capacitances
\(C_1=\dfrac{\epsilon_0A}{3d}\), \(C_2=\dfrac{2\epsilon_0A}{3d}\), \(C_3=\dfrac{3\epsilon_0A}{3d}\).

Step 3: Add in parallel
\(C=C_1+C_2+C_3=\dfrac{\epsilon_0A}{3d}(1+2+3)=\dfrac{2A\epsilon_0}{d}\)

Final Answer:
The equivalent capacitance is 2 A epsilon-naught over d. \[ \boxed{\text{(B) }\dfrac{2A\epsilon_{0}}{d}} \]