MHT CET · Physics · Capacitance
Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity of capacitor will [A = area of plate. \(t_1, t_2\) and \(t_3\) are thickness of dielectric slabs, \(k_1\), \(\mathrm{k}_2\) and \(\mathrm{k}_3\) are dielectric constant.]
- A \(\frac{\mathrm{A} \varepsilon_0}{\left[\frac{\mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3}{\mathrm{k}_1+\mathrm{k}_2+\mathrm{k}_3}\right]}\)
- B \(\frac{\mathrm{A} \varepsilon_0\left(\mathrm{k}_1 \mathrm{k}_2 \mathrm{k}_3\right)}{\mathrm{t}_1 \mathrm{t}_2 \mathrm{t}_3}\)
- C \(\mathrm{~A} \varepsilon_0\left[\frac{\mathrm{k}_1}{\mathrm{t}_1}+\frac{\mathrm{k}_2}{\mathrm{t}_2}+\frac{\mathrm{k}_3}{\mathrm{t}_3}\right]\)
- D \(\frac{\mathrm{A} \varepsilon_0}{\left[\frac{\mathrm{t}_1}{\mathrm{k}_1}+\frac{\mathrm{t}_2}{\mathrm{k}_2}+\frac{\mathrm{t}_3}{\mathrm{k}_3}\right]}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{A} \varepsilon_0}{\left[\frac{\mathrm{t}_1}{\mathrm{k}_1}+\frac{\mathrm{t}_2}{\mathrm{k}_2}+\frac{\mathrm{t}_3}{\mathrm{k}_3}\right]}\)
Step-by-step Solution
Detailed explanation
These are three capacitors connected in series.
\(\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}=\frac{\mathrm{t}_1}{\varepsilon_0 \mathrm{Ak}_1}\) \(+~\frac{\mathrm{t}_2}{\varepsilon_0 \mathrm{Ak}_2}+\frac{\mathrm{t}_3}{\varepsilon_0 \mathrm{Ak}_3}\)
\(\mathrm{C}=\frac{\mathrm{A} \varepsilon_0}{\left[\frac{\mathrm{t}_1}{\mathrm{k}_1}+\frac{\mathrm{t}_2}{\mathrm{k}_2}+\frac{\mathrm{t}_3}{\mathrm{k}_3}\right]}\)
\(\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}=\frac{\mathrm{t}_1}{\varepsilon_0 \mathrm{Ak}_1}\) \(+~\frac{\mathrm{t}_2}{\varepsilon_0 \mathrm{Ak}_2}+\frac{\mathrm{t}_3}{\varepsilon_0 \mathrm{Ak}_3}\)
\(\mathrm{C}=\frac{\mathrm{A} \varepsilon_0}{\left[\frac{\mathrm{t}_1}{\mathrm{k}_1}+\frac{\mathrm{t}_2}{\mathrm{k}_2}+\frac{\mathrm{t}_3}{\mathrm{k}_3}\right]}\)
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