TS EAMCET · Physics · Capacitance
The plates of a parallel plate capacitor are charged upto 200 volts. A dielectirc slab of thickness \(4 \mathrm{~mm}\) is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by \(3.2 \mathrm{~mm}\). The dielectric constant of the dielectric slab is
- A \(1\)
- B \(4\)
- C \(5\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{d^{\prime}-t+\frac{t}{K}} \\ & \Rightarrow \quad d=d^{\prime}-t+\frac{t}{K} \Rightarrow d^{\prime}-d=t\left(1-\frac{1}{K}\right) \end{aligned} \] Here, \(d^{\prime}-d=3.2 \mathrm{~mm}, t=4 \mathrm{~mm}\)…
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