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The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.
- a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.
- b)The total charge on the plate X will be 2Q
- c)The total charge on the plate Y will be zero
- d)The cell will supply CE2 amount of energy
Correct answer is option 'A,B,C,D'. Can you explain this answer?
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Verified Answer
The two plates X and Y of a parallel plate capacitor of capacitance C ...
As the emf of cell is ϵ=Q/C, charge of amount Q will flow from the positive terminal to the negative terminal of the cell through the capacitor.
As the Y is connected to the negative terminal of the cell so it is ground and the total charge on plate Y will be zero.
Here cell charge the capacitor plate X and its total charge =(Q+Q)=2Q
The energy will supply by cell is U=Cϵ2
As the Y is connected to the negative terminal of the cell so it is ground and the total charge on plate Y will be zero.
Here cell charge the capacitor plate X and its total charge =(Q+Q)=2Q
The energy will supply by cell is U=Cϵ2
Most Upvoted Answer
The two plates X and Y of a parallel plate capacitor of capacitance C ...
When X is joined to positive terminal and Y is joined to negative terminal then it is clear that charge Q moves from negative to positive terminal in opposite direction of current bcz current always flow from from positive to negative. So from Y terminal charge Q flow so total charge on X is 2Q. Hence no charge remain on Y plate. And total energy in this condition is =1/2*2Q*E bcz charge is 2Q and E represent potential. Hence 1/2*2*E*C*E (Q=EC) given, so E square *C hence, all option correct
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The two plates X and Y of a parallel plate capacitor of capacitance C ...
Four statements are given regarding a parallel plate capacitor connected to a cell of emf E = Q/C. Let's analyze each statement:
a) Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.
This statement is true. When the parallel plate capacitor is connected to the cell, the positive terminal of the cell is connected to plate X, and the negative terminal is connected to plate Y. Since the plates have equal and opposite charges of amount Q, a charge of amount Q will flow from the negative terminal to the positive terminal of the cell to neutralize the charges on the plates.
b) The total charge on plate X will be 2Q.
This statement is false. Initially, both plates X and Y are given a charge of amount Q each. When the capacitor is connected to the cell, the charges redistribute, but the total charge remains the same. Therefore, the total charge on plate X will still be Q, not 2Q.
c) The total charge on plate Y will be zero.
This statement is false. Similar to the previous statement, the total charge on plate Y will also be Q. The charges redistribute, but the total charge on each plate remains the same.
d) The cell will supply CE^2 amount of energy.
This statement is true. The energy stored in a capacitor can be calculated using the formula: E = (1/2)CV^2, where E is the energy, C is the capacitance, and V is the potential difference across the capacitor. In this case, the potential difference V is equal to the emf of the cell, which is E = Q/C. Substituting this value into the formula, we get E = (1/2)C(Q/C)^2 = (1/2)Q^2/C. Therefore, the cell will supply CE^2 amount of energy.
In summary, all four statements are true.
a) Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.
This statement is true. When the parallel plate capacitor is connected to the cell, the positive terminal of the cell is connected to plate X, and the negative terminal is connected to plate Y. Since the plates have equal and opposite charges of amount Q, a charge of amount Q will flow from the negative terminal to the positive terminal of the cell to neutralize the charges on the plates.
b) The total charge on plate X will be 2Q.
This statement is false. Initially, both plates X and Y are given a charge of amount Q each. When the capacitor is connected to the cell, the charges redistribute, but the total charge remains the same. Therefore, the total charge on plate X will still be Q, not 2Q.
c) The total charge on plate Y will be zero.
This statement is false. Similar to the previous statement, the total charge on plate Y will also be Q. The charges redistribute, but the total charge on each plate remains the same.
d) The cell will supply CE^2 amount of energy.
This statement is true. The energy stored in a capacitor can be calculated using the formula: E = (1/2)CV^2, where E is the energy, C is the capacitance, and V is the potential difference across the capacitor. In this case, the potential difference V is equal to the emf of the cell, which is E = Q/C. Substituting this value into the formula, we get E = (1/2)C(Q/C)^2 = (1/2)Q^2/C. Therefore, the cell will supply CE^2 amount of energy.
In summary, all four statements are true.
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The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer?
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The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer?.
The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer?.
Solutions for The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer?, a detailed solution for The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer? has been provided alongside types of The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The two plates X and Y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. X is now joined to the positive terminal and Y to the negative terminal of a cell of emf E = Q/C.a)Charge of amount Q will flow from the negative terminal to the positive terminal of the cell inside it.b)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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