SATHEE: Electrostatics 5 Question 22

Electrostatics 5 Question 22

24. A parallel plate capacitor is made of two circular plates separated by a distance of $5 mm$ and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \times 10^{4} V / m$ , the charge density of the positive plate will be close to

(a) $6 \times 10^{- 7} C / m^{2}$

(b) $3 \times 10^{- 7} C / m^{2}$

(d) $6 \times 10^{4} C / m^{2}$

(2014 Main)

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Answer:

Correct Answer: 24. (a)

Solution:

When free space between parallel plates of capacitor,

$E = \frac{σ}{ε_{0}}$

When dielectric is introduced between parallel plates of capacitor, $E^{'} = \frac{σ}{K ε_{0}}$

Electric field inside dielectric, $\frac{σ}{K ε_{0}} = 3 \times 10^{4}$

where, $K$ =dielectric constant of medium $= 2.2$

$ε_{0} =$ permitivity of free space $= 8.85 \times 10^{- 12}$

$\begin{aligned} \Rightarrow σ & = 2.2 \times 8.85 \times 10^{- 12} \times 3 \times 10^{4} \\ = 6.6 \times 8.85 \times 10^{- 8} = 5.841 \times 10^{- 7} \\ = 6 \times 10^{- 7} C / m^{2} \end{aligned}$

Electrostatics 5 Question 22

24. A parallel plate capacitor is made of two circular plates separated by a distance of $5 mm$ and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \times 10^{4} V / m$ , the charge density of the positive plate will be close to

Answer:

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Electrostatics 5 Question 22

24. A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3×104 V/m, the charge density of the positive plate will be close to

Answer:

Solution:

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24. A parallel plate capacitor is made of two circular plates separated by a distance of $5 mm$ and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \times 10^{4} V / m$ , the charge density of the positive plate will be close to