UPSC Exam > UPSC Questions > A parallel plate capacitor has its plates hor...
Start Learning for Free
A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find?
Most Upvoted Answer
A parallel plate capacitor has its plates horizontal with air occupyin...
1. Given Parameters:
- Voltage (V) applied to the capacitor
- Initial separation between plates (d₁)
- Final separation between plates in equilibrium (d₂)
- Mass of the lower plate (m)
- Cross-sectional area of each plate (A)
2. Initial Situation:
- Initially, the lower plate is at a distance d₁ from the upper plate.
- The lower plate is connected to a spring, which will exert a force on it.
- The capacitor is connected to an electric source, creating an electric field between the plates.
- The electric field will create a force on the lower plate, causing it to move.
3. Equilibrium Situation:
- In equilibrium, the forces due to the electric field and the spring will balance out.
- The force due to the electric field is given by F₁ = Q * E, where Q is the charge on the plates and E is the electric field strength.
- The force due to the spring is given by F₂ = k * (d₂ - d₁), where k is the spring constant.
- Setting F₁ = F₂, we can calculate the value of d₂ in equilibrium.
4. Calculations:
- The charge on the plates can be calculated using Q = C * V, where C is the capacitance of the capacitor.
- The capacitance of a parallel plate capacitor is given by C = ε₀ * A / d, where ε₀ is the permittivity of free space.
- Substituting the values, we can calculate the charge on the plates and then determine the value of d₂.
5. Final Answer:
- By solving the equations for forces in equilibrium, we can find the final separation between the plates (d₂) when the system reaches stability under the applied voltage.
- Voltage (V) applied to the capacitor
- Initial separation between plates (d₁)
- Final separation between plates in equilibrium (d₂)
- Mass of the lower plate (m)
- Cross-sectional area of each plate (A)
2. Initial Situation:
- Initially, the lower plate is at a distance d₁ from the upper plate.
- The lower plate is connected to a spring, which will exert a force on it.
- The capacitor is connected to an electric source, creating an electric field between the plates.
- The electric field will create a force on the lower plate, causing it to move.
3. Equilibrium Situation:
- In equilibrium, the forces due to the electric field and the spring will balance out.
- The force due to the electric field is given by F₁ = Q * E, where Q is the charge on the plates and E is the electric field strength.
- The force due to the spring is given by F₂ = k * (d₂ - d₁), where k is the spring constant.
- Setting F₁ = F₂, we can calculate the value of d₂ in equilibrium.
4. Calculations:
- The charge on the plates can be calculated using Q = C * V, where C is the capacitance of the capacitor.
- The capacitance of a parallel plate capacitor is given by C = ε₀ * A / d, where ε₀ is the permittivity of free space.
- Substituting the values, we can calculate the charge on the plates and then determine the value of d₂.
5. Final Answer:
- By solving the equations for forces in equilibrium, we can find the final separation between the plates (d₂) when the system reaches stability under the applied voltage.
Attention UPSC Students!
To make sure you are not studying endlessly, EduRev has designed UPSC study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in UPSC.
|
Explore Courses for UPSC exam
|
|
Similar UPSC Doubts
Top Courses for UPSCView all
A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find?
Question Description
A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find?.
A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find?.
Solutions for A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find? in English & in Hindi are available as part of our courses for UPSC.
Download more important topics, notes, lectures and mock test series for UPSC Exam by signing up for free.
Here you can find the meaning of A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find? defined & explained in the simplest way possible. Besides giving the explanation of
A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find?, a detailed solution for A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find? has been provided alongside types of A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find? theory, EduRev gives you an
ample number of questions to practice A parallel plate capacitor has its plates horizontal with air occupying the space between the plates. The upper plate is fixed with a rigid support and the lower one is connected to a spring as shown in the figure. The distance between the plates is d₁. Now the capacitor is connected with an electric source with the voltage V. The separation between the plates changes to a_{2} while in equilibrium. The mass of the lower plate is m and the cross-sectional area of each plate is A. Find? tests, examples and also practice UPSC tests.
|
|
Explore Courses for UPSC 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
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup