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A dielectric slab of dielectric constant k is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge Q on the surface if slab is given by a) Q = q - q/k b) Q = -q q/k c) Q = q[1/k 1] d) Q= -q[1 1/k]?
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A dielectric slab of dielectric constant k is placed between the plate...
Explanation:
When a dielectric slab of dielectric constant k is placed between the plates of a parallel plate capacitor carrying charge q, it induces charges on its surfaces. The induced charge Q on the surface of the slab is given by:
Q = q(k-1)
This can be derived using the concept of capacitance. The capacitance of a parallel plate capacitor with vacuum between the plates is given by:
C = εA/d
Where ε is the permittivity of vacuum, A is the area of the plates, and d is the distance between them. When a dielectric of constant k is inserted between the plates, the capacitance becomes:
C' = kεA/d
As the charge on the plates remains constant, the potential difference between the plates decreases. This results in an induced charge on the surface of the dielectric. Let σ be the surface charge density on the plates and σ' be the surface charge density on the surfaces of the dielectric. Then,
σ = q/A
σ' = -σ/k
The negative sign indicates that the induced charge is opposite in sign to the charge on the plates. The total induced charge on the dielectric is given by:
Q' = σ'A = -σ/k * A = -q/k
The charge on the plates is distributed between the plates and the surfaces of the dielectric. Therefore, the total charge on the plates and the dielectric is:
q' = q + Q' = q(1-1/k)
Answer:
The correct option is (c) Q = q[1/k 1].
When a dielectric slab of dielectric constant k is placed between the plates of a parallel plate capacitor carrying charge q, it induces charges on its surfaces. The induced charge Q on the surface of the slab is given by:
Q = q(k-1)
This can be derived using the concept of capacitance. The capacitance of a parallel plate capacitor with vacuum between the plates is given by:
C = εA/d
Where ε is the permittivity of vacuum, A is the area of the plates, and d is the distance between them. When a dielectric of constant k is inserted between the plates, the capacitance becomes:
C' = kεA/d
As the charge on the plates remains constant, the potential difference between the plates decreases. This results in an induced charge on the surface of the dielectric. Let σ be the surface charge density on the plates and σ' be the surface charge density on the surfaces of the dielectric. Then,
σ = q/A
σ' = -σ/k
The negative sign indicates that the induced charge is opposite in sign to the charge on the plates. The total induced charge on the dielectric is given by:
Q' = σ'A = -σ/k * A = -q/k
The charge on the plates is distributed between the plates and the surfaces of the dielectric. Therefore, the total charge on the plates and the dielectric is:
q' = q + Q' = q(1-1/k)
Answer:
The correct option is (c) Q = q[1/k 1].
Community Answer
A dielectric slab of dielectric constant k is placed between the plate...
The induced charge can be estimated as Qinduced = Qplate[K−1/ K]
Where K is the dielectric constant and K>1 for all the dielectric.
Where K is the dielectric constant and K>1 for all the dielectric.
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A dielectric slab of dielectric constant k is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge Q on the surface if slab is given by a) Q = q - q/k b) Q = -q q/k c) Q = q[1/k 1] d) Q= -q[1 1/k]?
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A dielectric slab of dielectric constant k is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge Q on the surface if slab is given by a) Q = q - q/k b) Q = -q q/k c) Q = q[1/k 1] d) Q= -q[1 1/k]? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A dielectric slab of dielectric constant k is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge Q on the surface if slab is given by a) Q = q - q/k b) Q = -q q/k c) Q = q[1/k 1] d) Q= -q[1 1/k]? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dielectric slab of dielectric constant k is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge Q on the surface if slab is given by a) Q = q - q/k b) Q = -q q/k c) Q = q[1/k 1] d) Q= -q[1 1/k]?.
A dielectric slab of dielectric constant k is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge Q on the surface if slab is given by a) Q = q - q/k b) Q = -q q/k c) Q = q[1/k 1] d) Q= -q[1 1/k]? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A dielectric slab of dielectric constant k is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge Q on the surface if slab is given by a) Q = q - q/k b) Q = -q q/k c) Q = q[1/k 1] d) Q= -q[1 1/k]? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dielectric slab of dielectric constant k is placed between the plates of a parallel plate capacitor carrying charge q. The induced charge Q on the surface if slab is given by a) Q = q - q/k b) Q = -q q/k c) Q = q[1/k 1] d) Q= -q[1 1/k]?.
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