SATHEE: Electrostatics 5 Question 7

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, $ε_{0} = 8.86 \times 10^{- 12} C^{2} / N - m^{2}$ )

(Main 2019, 8 April I)

(a) 3.8

(b) 8.5

(d) 6.2

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Answer:

Correct Answer: 8. (b)

Solution:

As we know, capacitance of a capacitor filled with dielectric medium,

$C = \frac{ε_{0} K A}{d}$

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

$K = \frac{C V}{ε_{0} A E}$

Given, $C = 15 pF = 15 \times 10^{- 12} F$ ,

$V = 500 V, E = 10^{6} {Vm}^{- 1},$

$A = 10^{- 4} m^{2}$

and $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$

Substituting the values in Eq. (iii), we get

$\begin{aligned} K & = \frac{15 \times 10^{- 12} \times 500}{8.85 \times 10^{- 12} \times 10^{- 4} \times 10^{6}} \\ = 8.47 \approx 8.5 \end{aligned}$

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 $15 pF$ ?

Answer:

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Electrostatics 5 Question 7

8. Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 106 V/m. The plate area is 10−4 m2. What is the dielectric constant, if the capacitance is 15pF ?

Answer:

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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 $15 pF$ ?