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A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.
(Round off to two decimal places)
(Round off to two decimal places)
Correct answer is '-0.25'. Can you explain this answer?
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Verified Answer
A parallel-plate capacitor is charged to a potential V0, charged Q0 an...
Initial charge on the plates and potential difference between the plates are Q0 and V0 respectively. So, the electrostatic energy is stored inside the capacitor is
Now, if we reduce tlie separation between two plates by a factor 1/2 keeping charge fixed, tlie potential difference two plates wil be
Tlierefore, the final electrostatic energy is stored inside the capacitor is,
Therefore, the charge of electrostatic energy;
Now, if we reduce tlie separation between two plates by a factor 1/2 keeping charge fixed, tlie potential difference two plates wil be
Tlierefore, the final electrostatic energy is stored inside the capacitor is,
Therefore, the charge of electrostatic energy;
Most Upvoted Answer
A parallel-plate capacitor is charged to a potential V0, charged Q0 an...
The Change in Electrostatic Energy of a Parallel-Plate Capacitor
Given Information:
- A parallel-plate capacitor is charged to a potential V0 and charge Q0.
- The capacitor is then disconnected from the battery.
- The separation of the plates is halved.
Explanation:
1. Definition of Electrostatic Energy:
- The electrostatic energy of a capacitor is given by the formula: U = (1/2)CV^2, where U is the energy, C is the capacitance, and V is the potential difference across the capacitor.
2. Capacitance of a Parallel-Plate Capacitor:
- The capacitance of a parallel-plate capacitor is given by the formula: C = εA/d, where ε is the permittivity of the medium between the plates, A is the area of each plate, and d is the separation between the plates.
3. Relationship between Charge, Potential Difference, and Capacitance:
- The charge on a capacitor can be expressed as: Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference across the capacitor.
4. Initial Energy of the Capacitor:
- The initial energy of the capacitor can be calculated using the formula: U0 = (1/2)C0V0^2, where U0 is the initial energy, C0 is the initial capacitance, and V0 is the initial potential difference.
5. Relationship between Capacitance and Separation:
- The capacitance of a parallel-plate capacitor is inversely proportional to the separation between the plates. Therefore, when the separation is halved, the capacitance doubles.
6. New Capacitance of the Capacitor:
- After the separation is halved, the new capacitance can be calculated using the formula: C' = εA/(d/2) = 2εA/d = 2C0.
7. New Potential Difference of the Capacitor:
- The charge on the capacitor remains constant, so the new potential difference can be calculated using the formula: V' = Q/C' = Q/(2C0) = V0/2.
8. New Energy of the Capacitor:
- The new energy of the capacitor can be calculated using the formula: U' = (1/2)C'V'^2 = (1/2)(2C0)(V0/2)^2 = (1/2)(C0)(V0^2/4) = (1/8)(C0)(V0^2) = (1/8)U0.
9. Change in Electrostatic Energy:
- The change in electrostatic energy can be calculated by subtracting the new energy from the initial energy: ΔU = U' - U0 = (1/8)U0 - U0 = -(7/8)U0 = -0.875U0.
10. Rounding Off the Answer:
- Rounding off the change in electrostatic energy to two decimal places gives -0.88U0.
Conclusion:
- The change in electrostatic energy, rounded off to two decimal places, is -0
Given Information:
- A parallel-plate capacitor is charged to a potential V0 and charge Q0.
- The capacitor is then disconnected from the battery.
- The separation of the plates is halved.
Explanation:
1. Definition of Electrostatic Energy:
- The electrostatic energy of a capacitor is given by the formula: U = (1/2)CV^2, where U is the energy, C is the capacitance, and V is the potential difference across the capacitor.
2. Capacitance of a Parallel-Plate Capacitor:
- The capacitance of a parallel-plate capacitor is given by the formula: C = εA/d, where ε is the permittivity of the medium between the plates, A is the area of each plate, and d is the separation between the plates.
3. Relationship between Charge, Potential Difference, and Capacitance:
- The charge on a capacitor can be expressed as: Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference across the capacitor.
4. Initial Energy of the Capacitor:
- The initial energy of the capacitor can be calculated using the formula: U0 = (1/2)C0V0^2, where U0 is the initial energy, C0 is the initial capacitance, and V0 is the initial potential difference.
5. Relationship between Capacitance and Separation:
- The capacitance of a parallel-plate capacitor is inversely proportional to the separation between the plates. Therefore, when the separation is halved, the capacitance doubles.
6. New Capacitance of the Capacitor:
- After the separation is halved, the new capacitance can be calculated using the formula: C' = εA/(d/2) = 2εA/d = 2C0.
7. New Potential Difference of the Capacitor:
- The charge on the capacitor remains constant, so the new potential difference can be calculated using the formula: V' = Q/C' = Q/(2C0) = V0/2.
8. New Energy of the Capacitor:
- The new energy of the capacitor can be calculated using the formula: U' = (1/2)C'V'^2 = (1/2)(2C0)(V0/2)^2 = (1/2)(C0)(V0^2/4) = (1/8)(C0)(V0^2) = (1/8)U0.
9. Change in Electrostatic Energy:
- The change in electrostatic energy can be calculated by subtracting the new energy from the initial energy: ΔU = U' - U0 = (1/8)U0 - U0 = -(7/8)U0 = -0.875U0.
10. Rounding Off the Answer:
- Rounding off the change in electrostatic energy to two decimal places gives -0.88U0.
Conclusion:
- The change in electrostatic energy, rounded off to two decimal places, is -0
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A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.(Round off to two decimal places)Correct answer is '-0.25'. Can you explain this answer?
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A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.(Round off to two decimal places)Correct answer is '-0.25'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.(Round off to two decimal places)Correct answer is '-0.25'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.(Round off to two decimal places)Correct answer is '-0.25'. Can you explain this answer?.
A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.(Round off to two decimal places)Correct answer is '-0.25'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.(Round off to two decimal places)Correct answer is '-0.25'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.(Round off to two decimal places)Correct answer is '-0.25'. Can you explain this answer?.
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A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.(Round off to two decimal places)Correct answer is '-0.25'. Can you explain this answer?, a detailed solution for A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.(Round off to two decimal places)Correct answer is '-0.25'. Can you explain this answer? has been provided alongside types of A parallel-plate capacitor is charged to a potential V0, charged Q0 and then disconnected from the battery. The separation of the plates is then halved. The change of electrostatic energy is ____________Q0V0.(Round off to two decimal places)Correct answer is '-0.25'. Can you explain this answer? theory, EduRev gives you an
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