Question

A parallel plate capacitor of area A and plate separation d is filled with two dielectrics as shown. What is the capacitance of the arrangement? 

• A. \(\dfrac{3K\varepsilon_0A}{4d}\)
• B. \(\dfrac{4K\varepsilon_0A}{3d}\)
• C. \(\dfrac{(K+1)\varepsilon_0A}{2d}\)
• D. (K(K+3)varepsilon₀A)/(2(K+1)d)

Hint:

• Same potential difference ⟹ parallel combination • Same charge ⟹ series combination Always identify geometry first.

Answer 1

The Correct Option is D

Step 1: Left half of area (A)/(2) has two dielectrics in series: C₁=((d/2)/(Kvarepsilon₀(A/2))+(d/2)/(varepsilon₀(A/2)))⁻1 =(Kvarepsilon₀A)/((K+1)d)
Step 2: Right half of area (A)/(2) filled with dielectric K: C₂=(Kvarepsilon₀(A/2))/(d)
Step 3: Both parts are in parallel: C=C₁+C₂ =(varepsilon₀A)/(d)((K)/(K+1)+(K)/(2))
Step 4: Simplifying: C=(K(K+3)varepsilon₀A)/(2(K+1)d)