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A charged conductor produces an electric field , of Intensity 103 V/m just out side its surface in vacuum. Then, it produces the electric field of intensity E just outside Its surface, when it is placed in a medium of dielectric constant 4. The value of E will be
- a)400 V/m
- b)450 V/m
- c)250 V/m
- d)150 V/m
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A charged conductor produces an electric field , of Intensity 103 V/m...
Given , electric field produced by charged conductor in vacuum,
E0 = 103 V/m
Electric field produced by charged conductor, when it is placed in medium of dielectrics (K = 4) is E = E0/K = 103/4 = 250V/m
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Most Upvoted Answer
A charged conductor produces an electric field , of Intensity 103 V/m...
Calculation of Electric Field
Given: Intensity of Electric Field outside surface in vacuum = 103 V/m
We know that the electric field just outside the surface of a charged conductor is given by:
E = σ/ε0
where σ is the surface charge density and ε0 is the permittivity of free space.
Let σ1 be the surface charge density of the conductor in vacuum.
Then, we have:
103 = σ1/ε0
⇒ σ1 = 103 × ε0
Calculation of Electric Field in Dielectric Medium
Given: Dielectric constant of the medium = 4
We know that when a charged conductor is placed in a dielectric medium, the electric field just outside its surface is given by:
E = σ/ε
where σ is the surface charge density and ε is the permittivity of the medium.
Let σ2 be the surface charge density of the conductor in the dielectric medium.
Then, we have:
E = σ2/ε
⇒ σ2 = E × ε
We also know that the permittivity of a medium is related to its dielectric constant by:
ε = ε0 × εr
where εr is the relative permittivity or dielectric constant of the medium.
Substituting the values, we get:
ε = ε0 × 4 = 4ε0
σ2 = E × ε = E × 4ε0
Comparing σ1 and σ2, we get:
σ2/σ1 = (E × 4ε0)/(103 × ε0) = 4E/103
Therefore, the electric field just outside the surface of the conductor in the dielectric medium is:
E = (σ2/ε) = (σ1/ε0) × (σ2/σ1) = 103 × (4E/103) = 4E
Hence, the value of E is 250 V/m (Option C).
Given: Intensity of Electric Field outside surface in vacuum = 103 V/m
We know that the electric field just outside the surface of a charged conductor is given by:
E = σ/ε0
where σ is the surface charge density and ε0 is the permittivity of free space.
Let σ1 be the surface charge density of the conductor in vacuum.
Then, we have:
103 = σ1/ε0
⇒ σ1 = 103 × ε0
Calculation of Electric Field in Dielectric Medium
Given: Dielectric constant of the medium = 4
We know that when a charged conductor is placed in a dielectric medium, the electric field just outside its surface is given by:
E = σ/ε
where σ is the surface charge density and ε is the permittivity of the medium.
Let σ2 be the surface charge density of the conductor in the dielectric medium.
Then, we have:
E = σ2/ε
⇒ σ2 = E × ε
We also know that the permittivity of a medium is related to its dielectric constant by:
ε = ε0 × εr
where εr is the relative permittivity or dielectric constant of the medium.
Substituting the values, we get:
ε = ε0 × 4 = 4ε0
σ2 = E × ε = E × 4ε0
Comparing σ1 and σ2, we get:
σ2/σ1 = (E × 4ε0)/(103 × ε0) = 4E/103
Therefore, the electric field just outside the surface of the conductor in the dielectric medium is:
E = (σ2/ε) = (σ1/ε0) × (σ2/σ1) = 103 × (4E/103) = 4E
Hence, the value of E is 250 V/m (Option C).
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A charged conductor produces an electric field , of Intensity 103 V/m just out side its surface in vacuum. Then, it produces the electric field of intensity E just outside Its surface, when it is placed in a medium of dielectric constant 4. The value of E will be a)400 V/mb)450 V/mc)250 V/md)150 V/mCorrect answer is option 'C'. Can you explain this answer?
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A charged conductor produces an electric field , of Intensity 103 V/m just out side its surface in vacuum. Then, it produces the electric field of intensity E just outside Its surface, when it is placed in a medium of dielectric constant 4. The value of E will be a)400 V/mb)450 V/mc)250 V/md)150 V/mCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A charged conductor produces an electric field , of Intensity 103 V/m just out side its surface in vacuum. Then, it produces the electric field of intensity E just outside Its surface, when it is placed in a medium of dielectric constant 4. The value of E will be a)400 V/mb)450 V/mc)250 V/md)150 V/mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A charged conductor produces an electric field , of Intensity 103 V/m just out side its surface in vacuum. Then, it produces the electric field of intensity E just outside Its surface, when it is placed in a medium of dielectric constant 4. The value of E will be a)400 V/mb)450 V/mc)250 V/md)150 V/mCorrect answer is option 'C'. Can you explain this answer?.
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