NEET · Physics · STD 12 - 2. Electric potential and capacitance
The plates of a parallel plate capacitor are separated by \(d\). Two slabs of different dielectric constant \(K_1\) and \(K_2\) with thickness \(\frac{3}{8} d\) and \(\frac{d}{2}\), respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates.
(If \(K_1=1.25 K_2\), the value of \(K_1\) is:
- A 2.66
- B 2.33
- C 1.6
- D 1.33
Answer & Solution
Correct Answer
(A) 2.66
Step-by-step Solution
Detailed explanation
\(C_1=\varepsilon_0 A / d\)
\(C _2=\frac{\varepsilon_0 A}{\frac{3 d}{8 K_1}+\frac{ d }{2 K_2}+\left( d -\frac{3 d}{8}-\frac{ d }{2}\right)} \)
\( C _2=\frac{\varepsilon_0 A / d }{\frac{3}{8 K_1}+\frac{1}{2 K_2}+\frac{1}{8}} \)
\( =\frac{\varepsilon_0 A / d }{\frac{3}{8 K_1}+\frac{1}{2\left(\frac{4}{5} K_1\right)}+\frac{1}{8}} \)
\( C _2=2 C _1 \)
\( \frac{\varepsilon_0 A / d }{\frac{3}{8 K_1}+\frac{5}{8 K_1}+\frac{1}{8}}=2 \varepsilon_0 A / d \)
\( \Rightarrow \frac{1}{\frac{1}{K_1}+\frac{1}{8}}=2 \Rightarrow \frac{1}{K_1}+\frac{1}{8}=\frac{1}{2} \Rightarrow K_1=\frac{8}{3}=2.66\)
\(C _2=\frac{\varepsilon_0 A}{\frac{3 d}{8 K_1}+\frac{ d }{2 K_2}+\left( d -\frac{3 d}{8}-\frac{ d }{2}\right)} \)
\( C _2=\frac{\varepsilon_0 A / d }{\frac{3}{8 K_1}+\frac{1}{2 K_2}+\frac{1}{8}} \)
\( =\frac{\varepsilon_0 A / d }{\frac{3}{8 K_1}+\frac{1}{2\left(\frac{4}{5} K_1\right)}+\frac{1}{8}} \)
\( C _2=2 C _1 \)
\( \frac{\varepsilon_0 A / d }{\frac{3}{8 K_1}+\frac{5}{8 K_1}+\frac{1}{8}}=2 \varepsilon_0 A / d \)
\( \Rightarrow \frac{1}{\frac{1}{K_1}+\frac{1}{8}}=2 \Rightarrow \frac{1}{K_1}+\frac{1}{8}=\frac{1}{2} \Rightarrow K_1=\frac{8}{3}=2.66\)
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