Four plates of the same area of cross-section are joined as shown in the figure. The distance between each plate is $d$. The equivalent capacity across $A$ and $B$ will be
  • A$\frac{{2{\varepsilon _0}A}}{d}$
  • B$\frac{{3{\varepsilon _0}A}}{d}$
  • C$\frac{{3{\varepsilon _0}A}}{{2d}}$
  • D$\frac{{{\varepsilon _0}A}}{d}$
Medium
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