SATHEE: Electrostatics Question 103

Electrostatics Question 103

Question: In a capacitor of capacitance $20 μ F$ , the distance between the plates is 2mm. If a dielectric slab of width 1mm and dielectric constant 2 is inserted between the plates, then the new capacitance is [BHU 2000]

Options:

A) $2 μ F$

B) $15.5 μ F$

C) $26.6 μ F$

D) $32 μ F$

Show Answer

Answer:

Correct Answer: C

Solution:

$C = \frac{ε_{0} A}{d}$ and $C^{'} = \frac{ε_{0} A}{(d - t + \frac{t}{K})}$

therefore $\frac{C}{C^{'}} = \frac{(d - t + \frac{t}{K})}{d}$

$\Rightarrow \frac{20}{C^{'}} = \frac{(2 \times 10^{- 3} - 1 \times 10^{- 3} + \frac{1 \times 10^{- 3}}{2})}{2 \times 10^{- 3}}$

therefore $C^{'} = 26.6 μ F$

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Electrostatics Question 103

Question: In a capacitor of capacitance $20 μ F$ , the distance between the plates is 2mm. If a dielectric slab of width 1mm and dielectric constant 2 is inserted between the plates, then the new capacitance is [BHU 2000]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

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अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Electrostatics Question 103

Question: In a capacitor of capacitance 20μF , the distance between the plates is 2mm. If a dielectric slab of width 1mm and dielectric constant 2 is inserted between the plates, then the new capacitance is [BHU 2000]

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: In a capacitor of capacitance $20 μ F$ , the distance between the plates is 2mm. If a dielectric slab of width 1mm and dielectric constant 2 is inserted between the plates, then the new capacitance is [BHU 2000]