Electrostatics 5 Question 7
8. Voltage rating of a parallel plate capacitor is $500 V$. Its dielectric can withstand a maximum electric field of $10^{6} V / m$. The plate area is $10^{-4} m^{2}$. What is the dielectric constant, if the capacitance is $15 pF$ ?
(Take, $\varepsilon _0=8.86 \times 10^{-12} C^{2} / N-m^{2}$ )
(Main 2019, 8 April I)
(a) 3.8
(b) 8.5
(c) 4.5
(d) 6.2
Show Answer
Answer:
Correct Answer: 8. $Q _0=\frac{C V R _2}{R _1+R _2}, \alpha=\frac{R _1+R _2}{C R _1 R _2}$ 9. (a) $Q=\frac{C V}{2}\left(1-e^{-2 t / 3 R C}\right)$
(b) $i _2=\frac{V}{2 R}-\frac{V}{6 R} e^{-2 t / 3 R C}, \frac{V}{2 R}$
Solution:
- As we know, capacitance of a capacitor filled with dielectric medium,
$$ C=\frac{\varepsilon _0 K A}{d} $$
and potential difference between plates is
$$ E=\frac{V}{d} \Rightarrow d=\frac{V}{E} $$
So, by combining both Eqs. (i) and (ii), we get
$$ K=\frac{C V}{\varepsilon _0 A E} $$
Given, $C=15 pF=15 \times 10^{-12} F$,
$$ V=500 V, E=10^{6} Vm^{-1}, $$
$$ A=10^{-4} m^{2} $$
and $\quad \varepsilon _0=8.85 \times 10^{-12} C^{2} N^{-1} m^{-2}$
Substituting the values in Eq. (iii), we get
$$ \begin{aligned} K & =\frac{15 \times 10^{-12} \times 500}{8.85 \times 10^{-12} \times 10^{-4} \times 10^{6}} \\ & =8.47 \approx 8.5 \end{aligned} $$
