SATHEE: Electrostatics Question 85

Electrostatics Question 85

Question: The capacity of a parallel plate condenser is $10 μ F$ without dielectric. Dielectric of constant 2 is used to fill half the distance between the plates, the new capacitance in $μ F$ is [EAMCET (Engg.) 1995]

Options:

A) 10

B) 20

C) 15

D) 13.33

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Answer:

Correct Answer: D

Solution:

$C = \frac{A ε_{0}}{d} = 10 μ F$

$C_{1} = \frac{A ε_{0}}{d - t + \frac{t}{k}} = \frac{A ε_{0}}{d - \frac{d}{2} + \frac{d}{2 k}} = \frac{A ε_{0}}{\frac{d}{2} (1 + \frac{1}{2})} = \frac{4}{3} . \frac{A ε_{0}}{d}$

$C_{1} = \frac{4}{3} \times 10 = 13.33 μ F$

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Electrostatics Question 85

Question: The capacity of a parallel plate condenser is $10 μ F$ without dielectric. Dielectric of constant 2 is used to fill half the distance between the plates, the new capacitance in $μ F$ is [EAMCET (Engg.) 1995]

Options:

Answer:

Solution:

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Electrostatics Question 85

Question: The capacity of a parallel plate condenser is 10μF without dielectric. Dielectric of constant 2 is used to fill half the distance between the plates, the new capacitance in μF is [EAMCET (Engg.) 1995]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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Question: The capacity of a parallel plate condenser is $10 μ F$ without dielectric. Dielectric of constant 2 is used to fill half the distance between the plates, the new capacitance in $μ F$ is [EAMCET (Engg.) 1995]