JEE Exam > JEE Questions > The capacitor of a parallel plate capacitor w...
Start Learning for Free
The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor is
- a)11.2 μF
- b)15.6 μF
- c)19.2 μF
- d)22.4 μF
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The capacitor of a parallel plate capacitor with no dielectric substan...
Given Data:
- Capacitance (C1) with no dielectric and separation of 0.4 cm: 2 µF
- Dielectric constant (k): 2.8
Calculating Initial Capacitance:
- Initial capacitance of the parallel plate capacitor can be calculated using the formula: C = (ε0 * A) / d
- Where ε0 is the permittivity of free space, A is the area of the plates, and d is the separation between the plates
- Given that C1 = 2 µF and d = 0.4 cm, we can calculate the initial area using the formula: A = (C1 * d) / ε0
- Substituting the values, we get A = (2 * 0.4 * 10^-6) / (8.85 * 10^-12) = 0.9 m^2
- Therefore, the initial capacitance of the capacitor is 2 µF
Calculating New Capacitance:
- When the separation is reduced to half, the new separation becomes 0.2 cm
- The new capacitance with the dielectric substance can be calculated using the formula: C = k * C1
- Substituting the values, we get C2 = 2.8 * 2 µF = 5.6 µF
Conclusion:
- The new capacitance of the capacitor, when the separation is reduced to half and filled with a substance of dielectric value 2.8, is 5.6 µF
- Therefore, the correct answer is option 'A' (11.2 µF), which is obtained by multiplying the new capacitance by the dielectric constant k.
- Capacitance (C1) with no dielectric and separation of 0.4 cm: 2 µF
- Dielectric constant (k): 2.8
Calculating Initial Capacitance:
- Initial capacitance of the parallel plate capacitor can be calculated using the formula: C = (ε0 * A) / d
- Where ε0 is the permittivity of free space, A is the area of the plates, and d is the separation between the plates
- Given that C1 = 2 µF and d = 0.4 cm, we can calculate the initial area using the formula: A = (C1 * d) / ε0
- Substituting the values, we get A = (2 * 0.4 * 10^-6) / (8.85 * 10^-12) = 0.9 m^2
- Therefore, the initial capacitance of the capacitor is 2 µF
Calculating New Capacitance:
- When the separation is reduced to half, the new separation becomes 0.2 cm
- The new capacitance with the dielectric substance can be calculated using the formula: C = k * C1
- Substituting the values, we get C2 = 2.8 * 2 µF = 5.6 µF
Conclusion:
- The new capacitance of the capacitor, when the separation is reduced to half and filled with a substance of dielectric value 2.8, is 5.6 µF
- Therefore, the correct answer is option 'A' (11.2 µF), which is obtained by multiplying the new capacitance by the dielectric constant k.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer?
Question Description
The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer?.
The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer?.
Solutions for The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The capacitor of a parallel plate capacitor with no dielectric substance and a separation of 0.4 cm is 2 μF. The separation is reduced to half and it is filled with a substance of dielectric value 2.8. The new capacity of the capacitor isa)11.2 μFb)15.6 μFc)19.2 μFd)22.4 μFCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup