MHT CET · Physics · Capacitance
In a parallel plate air capacitor of plate separation \(d\), a dielectric slab of thickness \(t\) is introduced between the plates \((t < d)\). The capacitance becomes one-third of the original value. The dielectric constant of the slab will be
- A \(\frac{t}{2 d+t}\)
- B \(\frac{t}{d-2 t}\)
- C \(\frac{t}{d+t}\)
- D \(\frac{2 t}{2 d-t}\)
Answer & Solution
Correct Answer
(A) \(\frac{t}{2 d+t}\)
Step-by-step Solution
Detailed explanation
Equivalent Capacitance of the dielectric-air system is given by
\(\frac{1}{C^{\prime}}=\left(\frac{d-t}{\varepsilon_0 A}\right)+\left(\frac{t}{K \varepsilon_0 A}\right)=\frac{d-t\left(1-\frac{1}{K}\right)}{\varepsilon_0 A}\)
If new capacitance is given by one third of the old value:
\(\begin{aligned}
& C=\frac{\varepsilon_0 A}{d-t\left(1-\frac{1}{K}\right)}=\frac{\varepsilon_0 A}{3 d} \\
& \Rightarrow d-t+\frac{t}{K}=3 d \\
& \Rightarrow K=\frac{t}{2 d+t}
\end{aligned}\)
\(\frac{1}{C^{\prime}}=\left(\frac{d-t}{\varepsilon_0 A}\right)+\left(\frac{t}{K \varepsilon_0 A}\right)=\frac{d-t\left(1-\frac{1}{K}\right)}{\varepsilon_0 A}\)
If new capacitance is given by one third of the old value:
\(\begin{aligned}
& C=\frac{\varepsilon_0 A}{d-t\left(1-\frac{1}{K}\right)}=\frac{\varepsilon_0 A}{3 d} \\
& \Rightarrow d-t+\frac{t}{K}=3 d \\
& \Rightarrow K=\frac{t}{2 d+t}
\end{aligned}\)
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