SATHEE: Electrostatics Question 58

Electrostatics Question 58

Question: A parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to potential $V$ and then the battery is disconnected. A slab of dielectric constant $k$ is then inserted between the plates of the capacitors so as to fill the space between the plates. If $Q, E$ and $W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and work done on the system in question in the process of inserting the slab, then state incorrect relation from the following [IIT 1991; MP PET 1997]

Options:

A) $Q = \frac{ε_{0} A V}{d}$

B) $W = \frac{ε_{0} A V^{2}}{2 k d}$

C) $E = \frac{V}{k d}$

D) $W = \frac{ε_{0} A V^{2}}{2 d} (1 - \frac{1}{k})$

Show Answer

Answer:

Correct Answer: B

Solution:

After inserting the dielectric slab New capacitance $C^{'} = K . C = \frac{K ε_{0} A}{d}$

New potential difference $V^{'} = \frac{V}{K}$ New charge $Q^{'} = C^{'} V^{'} = \frac{ε_{0} A V}{d}$

New electric field $E^{'} = \frac{V^{'}}{d} = \frac{V}{K d}$ Work done (W) = Final energy - Initial energy W $= \frac{1}{2} C^{'} V {^{'}}^{2} - \frac{1}{2} C V^{2}$

$= \frac{1}{2} (K C) {(\frac{V}{K})}^{2} - \frac{1}{2} C V^{2}$

$= \frac{1}{2} C V^{2} (\frac{1}{K} - 1) = - \frac{1}{2} C V^{2} (1 - \frac{1}{K})$

$= - \frac{ε_{0} A V^{2}}{2 d} (1 - \frac{1}{K})$ so $| W | = \frac{ε_{0} A V^{2}}{2 d} (1 - \frac{1}{K})$ .

पिछला

अगला

Electrostatics Question 58

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

Electrostatics Question 58

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

Menu