Question

When x = \(\frac{d}{3}\) then capacitance is C₁ = 2 μF then if x = \(\frac{2d}{3}\) then capacitance C₂ will be (in μF).

Answer 1

Correct Answer: 3

\(\frac{1}{C_{eq}}=\frac{1}{C_{air}}+\frac{1}{C_{di}}\)
\(\frac{1}{C_{eq}}=\frac{2d}{3(\epsilon _{0}A)}+\frac{d}{(3)4\epsilon _{0}A}=\frac{3d}{4A\epsilon _{0}}\)
\(C_{eq}=\frac{4A\epsilon _{0}}{3d}=2\mu F\)
\(\frac{A\epsilon _{0}}{d}=1.5\mu F\)
\(\frac{1}{C'_{eq}}=\frac{d}{3(\epsilon _{0}A)}+\frac{2d}{(3)4\epsilon _{0}A}\)= \(\frac{(4+2)d}{12\epsilon _{0}{A}}\)\(=\frac{6d}{12\epsilon _{0}{A}}\)\(\Rightarrow C'_{eq}=2[\frac{\epsilon _{0}A}{d}]=3\mu F\)

So, the correct answer is 3 μF