Question Description
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 ib)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2 amount 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 ib)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2 amount 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 ib)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2 amount 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 ib)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2 amount 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. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.

Here you can find the meaning 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 ib)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2 amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation 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 ib)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2 amount 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 ib)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2 amount 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 ib)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2 amount 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 ib)The total charge on the plate X will be 2Qc)The total charge on the plate Y will be zerod)The cell will supply CE2 amount of energyCorrect answer is option 'A,B,C,D'. Can you explain this answer? tests, examples and also practice Class 12 tests.