AP EAMCET · PHYSICS · Capacitance
Four metal plates, each with surface area \(A\) on one side, are placed with separation ' \(\mathrm{d}\) ' as shown in the figure. The capacitance between a and \(b\) is
\(\left(\varepsilon_0-\right.\) permittivity of free space)
- A \(\frac{3 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
- B \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
- C \(\frac{2 \varepsilon_0 \mathrm{~A}}{3 \mathrm{~d}}\)
- D \(\frac{3 \varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}\)
Answer & Solution
Correct Answer
(D) \(\frac{3 \varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{C}_1 \equiv \mathrm{C}_2=\mathrm{C}_3=\mathrm{C}\) Capacitor \(\mathrm{C}_1 \& \mathrm{C}_2\) are in series. \(C_{12}=\frac{C_1 C_2}{C_1+C_2}=\frac{C^2}{2 C}=\frac{C}{2}\) Now \(\mathrm{C}_{12}\) \& \(\mathrm{C}_3\) are in parallel,…
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