Q161 / 180 · KCET 2014 · Physics
A voltmeter reads \( 4 \mathrm{~V} \) when connected to a parallel plate capacitor with air as a dielectric.
When a dielectric slab is introduced between plates for the same configuration, voltmeter
reads \( 2 \mathrm{~V} \). What is the dielectric constant of the material ?
When a dielectric slab is introduced between plates for the same configuration, voltmeter
reads \( 2 \mathrm{~V} \). What is the dielectric constant of the material ?
- A\( 0.5 \)
- Correct answerB\( 12 \)
- C\( 08 \)
- D\( 10 \)
Solution
We know that \( Q=C V \)
\[ \begin{array}{l} \Rightarrow V=\frac{Q}{C} \\ \Rightarrow V \propto \frac{1}{C} \end{array} \]
Also \( C_{\text {dielectric }}=K C_{0} \)
Therefore, \( V_{\text {dielectric }} \propto \frac{1}{C_{\text {dielectric }}} \) and \( V_{0} \propto \frac{1}{C_{0}} \) and
\( \frac{V_{\text {dicloctic }}}{V_{0}}=\frac{C_{0}}{C_{\text {dielectric }}} \Rightarrow \frac{V_{\text {diclectric }}}{V_{0}}=\frac{C_{0}}{K C_{0}} \Rightarrow \frac{V_{\text {dielectric }}}{V_{0}}=\frac{1}{K} \)
Given \( V_{\text {diclectric }}=2 V ; V_{0}=4 V \)
\( \Rightarrow \frac{2}{4}=\frac{1}{K} \Rightarrow K=2 \)
Therefore, dielectric constant of the material is \( 2 \).
\[ \begin{array}{l} \Rightarrow V=\frac{Q}{C} \\ \Rightarrow V \propto \frac{1}{C} \end{array} \]
Also \( C_{\text {dielectric }}=K C_{0} \)
Therefore, \( V_{\text {dielectric }} \propto \frac{1}{C_{\text {dielectric }}} \) and \( V_{0} \propto \frac{1}{C_{0}} \) and
\( \frac{V_{\text {dicloctic }}}{V_{0}}=\frac{C_{0}}{C_{\text {dielectric }}} \Rightarrow \frac{V_{\text {diclectric }}}{V_{0}}=\frac{C_{0}}{K C_{0}} \Rightarrow \frac{V_{\text {dielectric }}}{V_{0}}=\frac{1}{K} \)
Given \( V_{\text {diclectric }}=2 V ; V_{0}=4 V \)
\( \Rightarrow \frac{2}{4}=\frac{1}{K} \Rightarrow K=2 \)
Therefore, dielectric constant of the material is \( 2 \).
Hence, the correct answer is option: 'B'.