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EEJ MAIN Physics QUESTION #6991
Question 1

A parallel plate capacitor (square plates of side $a$, separation $d$, with $d \ll a$) has its lower triangular half filled with a dielectric of constant $K$. What is the capacitance of this arrangement?

  • $\dfrac{K\varepsilon_0 a^2}{2d(K+1)}\ln K$
  • $\dfrac{K\varepsilon_0 a^2}{d(K-1)}\ln K$✔️
  • $\dfrac{K\varepsilon_0 a^2}{d}\ln K$
  • $\dfrac{1}{2}\dfrac{K\varepsilon_0 a^2}{d}$
Correct Answer Explanation

Consider the capacitor as strips of width $dx$ parallel to the shorter side. At horizontal position $x$ from the left, the gap has dielectric filling of height $\frac{d \cdot x}{a}$ (triangular region). The local gap (from bottom) filled with dielectric is $\frac{xd}{a}$ and air gap is $d - \frac{xd}{a} = \frac{d(a-x)}{a}$.

Each strip is two capacitors in series (dielectric + air). Integrating across the width gives:

$C = \int_0^a \frac{\varepsilon_0 K \cdot a \, dx}{d[(K-1)x/a + 1] \cdot a} \Rightarrow C = \frac{K\varepsilon_0 a^2}{d(K-1)}\ln K$