MHT CET · Physics · Capacitance
Air capacitor has capacitance ' \(\mathrm{C}_1\) '. The space between two plates of capacitor is filled with two dielectrics as shown in figure. The new capacitance of the capacitor is ' \(\mathrm{C}_2\) '. The ratio \(\frac{\mathrm{C}_1}{\mathrm{C}_2}\) is \((\mathrm{d}=\) distance between two plates of capacitor, \(K_1\) and \(K_2\) are dielectric constants of two dielectrics respectively)

- A \(\mathrm{K}_1+\mathrm{K}_2\)
- B \(\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1-\mathrm{K}_2}\)
- C \(\frac{2 \mathrm{~K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}\)
- D \(\frac{\mathrm{K}_1+\mathrm{K}_2}{2 \mathrm{~K}_1 \mathrm{~K}_2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{K}_1+\mathrm{K}_2}{2 \mathrm{~K}_1 \mathrm{~K}_2}\)
Step-by-step Solution
Detailed explanation
For an air capacitor, \(\mathrm{C}_1=\frac{\mathrm{A} \varepsilon_0}{\mathrm{~d}}\)
When two dielectrics are inserted as shown, the combination acts as capacitors in series.
\(\frac{1}{\mathrm{C}_2}=\frac{1}{\mathrm{CK}_1}+\frac{1}{\mathrm{CK}_2} \)
\( \frac{1}{\mathrm{C}_2}=\frac{1}{\frac{\mathrm{~A} \varepsilon_0 \mathrm{~K}_1}{\frac{\mathrm{~d}}{2}}}+\frac{1}{\frac{\mathrm{~A} \varepsilon_0 \mathrm{~K}_2}{\frac{d}{2}}}=\frac{\mathrm{d}}{2 \varepsilon_0 \mathrm{~A}}~+\) \(\left(\frac{1}{\mathrm{~K}_1}+\frac{1}{\mathrm{~K}_2}\right)\)
\(C_2 =\frac{2 \varepsilon_0 A}{d}+\left(\frac{K_1 K_2}{K_1+K_2}\right)=\frac{\frac{A \varepsilon_0}{d}}{\frac{2 A \varepsilon_0 A}{d}\left(\frac{K_1 K_2}{K_1+K_2}\right)} \)
\( =\frac{K_1+K_2}{2 K_1 K_2}\)
When two dielectrics are inserted as shown, the combination acts as capacitors in series.
\(\frac{1}{\mathrm{C}_2}=\frac{1}{\mathrm{CK}_1}+\frac{1}{\mathrm{CK}_2} \)
\( \frac{1}{\mathrm{C}_2}=\frac{1}{\frac{\mathrm{~A} \varepsilon_0 \mathrm{~K}_1}{\frac{\mathrm{~d}}{2}}}+\frac{1}{\frac{\mathrm{~A} \varepsilon_0 \mathrm{~K}_2}{\frac{d}{2}}}=\frac{\mathrm{d}}{2 \varepsilon_0 \mathrm{~A}}~+\) \(\left(\frac{1}{\mathrm{~K}_1}+\frac{1}{\mathrm{~K}_2}\right)\)
\(C_2 =\frac{2 \varepsilon_0 A}{d}+\left(\frac{K_1 K_2}{K_1+K_2}\right)=\frac{\frac{A \varepsilon_0}{d}}{\frac{2 A \varepsilon_0 A}{d}\left(\frac{K_1 K_2}{K_1+K_2}\right)} \)
\( =\frac{K_1+K_2}{2 K_1 K_2}\)
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