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A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.
Correct answer is '6'. Can you explain this answer?
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Most Upvoted Answer
A 60 pF capacitor is fully charged by a 20 V supply. It is then discon...
H = Ui - Uf =
= 1/4 × 60 × 10-12 × 400 × 109 nJ
= 6 nJ
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A 60 pF capacitor is fully charged by a 20 V supply. It is then discon...
Given:
- Capacitance of each capacitor, C = 60 pF = 60 × 10^(-12) F
- Initial voltage across the capacitor, V = 20 V
To Find:
- Electrostatic energy lost in the process
Formula:
- The energy stored in a capacitor is given by the formula: E = (1/2)CV^2
Explanation:
When the first capacitor is fully charged, it stores energy given by E1 = (1/2)C1V^2, where C1 is the capacitance of the first capacitor.
Step 1: Find the Energy Stored in the First Capacitor:
Given that C1 = 60 pF and V = 20 V, we can calculate the energy stored in the first capacitor as follows:
E1 = (1/2)(60 × 10^(-12) F)(20 V)^2
= (1/2)(60 × 10^(-12) F)(400 V^2)
= 12 × 10^(-12) F × 400 V^2
= 4800 × 10^(-12) V^2
= 4800 pJ
Step 2: Find the Total Energy Initially:
Since the second capacitor is uncharged initially, it does not store any energy. Therefore, the total energy initially is equal to the energy stored in the first capacitor:
Total initial energy = E1 = 4800 pJ
Step 3: Find the Energy Stored in the Second Capacitor:
When the two capacitors are connected in parallel, they share the charge and the voltage. The total charge Q is distributed between the two capacitors equally because they have the same capacitance. Therefore, each capacitor will have Q/2 charge.
The voltage across each capacitor is given by V = Q/C, where C is the capacitance of each capacitor. Therefore, the voltage across each capacitor is V = (Q/2)/(60 × 10^(-12) F) = Q/(120 × 10^(-12) F).
The energy stored in the second capacitor is given by E2 = (1/2)C2V^2, where C2 is the capacitance of the second capacitor.
E2 = (1/2)(60 × 10^(-12) F)(Q/(120 × 10^(-12) F))^2
= (1/2)(60/120)^2 Q^2
= (1/2)(1/4) Q^2
= (1/8) Q^2
Step 4: Find the Energy Lost:
The energy lost in the process is equal to the initial energy minus the energy stored in the second capacitor:
Energy lost = Total initial energy - Energy stored in the second capacitor
= E1 - E2
= 4800 pJ - (1/8) Q^2
Since the charge Q is distributed equally between the two capacitors, we can express Q in terms of the initial voltage V:
Q = CV
= (60 × 10^(-12) F)(20 V)
= 1200 × 10^(-12) C
- Capacitance of each capacitor, C = 60 pF = 60 × 10^(-12) F
- Initial voltage across the capacitor, V = 20 V
To Find:
- Electrostatic energy lost in the process
Formula:
- The energy stored in a capacitor is given by the formula: E = (1/2)CV^2
Explanation:
When the first capacitor is fully charged, it stores energy given by E1 = (1/2)C1V^2, where C1 is the capacitance of the first capacitor.
Step 1: Find the Energy Stored in the First Capacitor:
Given that C1 = 60 pF and V = 20 V, we can calculate the energy stored in the first capacitor as follows:
E1 = (1/2)(60 × 10^(-12) F)(20 V)^2
= (1/2)(60 × 10^(-12) F)(400 V^2)
= 12 × 10^(-12) F × 400 V^2
= 4800 × 10^(-12) V^2
= 4800 pJ
Step 2: Find the Total Energy Initially:
Since the second capacitor is uncharged initially, it does not store any energy. Therefore, the total energy initially is equal to the energy stored in the first capacitor:
Total initial energy = E1 = 4800 pJ
Step 3: Find the Energy Stored in the Second Capacitor:
When the two capacitors are connected in parallel, they share the charge and the voltage. The total charge Q is distributed between the two capacitors equally because they have the same capacitance. Therefore, each capacitor will have Q/2 charge.
The voltage across each capacitor is given by V = Q/C, where C is the capacitance of each capacitor. Therefore, the voltage across each capacitor is V = (Q/2)/(60 × 10^(-12) F) = Q/(120 × 10^(-12) F).
The energy stored in the second capacitor is given by E2 = (1/2)C2V^2, where C2 is the capacitance of the second capacitor.
E2 = (1/2)(60 × 10^(-12) F)(Q/(120 × 10^(-12) F))^2
= (1/2)(60/120)^2 Q^2
= (1/2)(1/4) Q^2
= (1/8) Q^2
Step 4: Find the Energy Lost:
The energy lost in the process is equal to the initial energy minus the energy stored in the second capacitor:
Energy lost = Total initial energy - Energy stored in the second capacitor
= E1 - E2
= 4800 pJ - (1/8) Q^2
Since the charge Q is distributed equally between the two capacitors, we can express Q in terms of the initial voltage V:
Q = CV
= (60 × 10^(-12) F)(20 V)
= 1200 × 10^(-12) C
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A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.Correct answer is '6'. Can you explain this answer?
Question Description
A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.Correct answer is '6'. Can you explain this answer?.
A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.Correct answer is '6'. Can you explain this answer?.
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A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.Correct answer is '6'. Can you explain this answer?, a detailed solution for A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.Correct answer is '6'. Can you explain this answer? has been provided alongside types of A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ________.Correct answer is '6'. Can you explain this answer? theory, EduRev gives you an
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