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The electric field on the surface of a perfect conductor is 2V/m. The conductor is immersed in water with ε = 80 ε0. The surface charge density on the conductor is ((ε0 = 10-9/36
p
)F/m)- a)0 c/m2
- b)2 c/m2
- c)1.8 x 10-11 c/m2
- d)1.14 x 10-9 c/m2
Correct answer is option 'D'. Can you explain this answer?
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Given data:
Electric field on the surface of a perfect conductor = 2 V/m
Dielectric constant of water (ε) = 80ε0
Where, ε0 = 8.85 x 10^-12 F/m
To find: Surface charge density on the conductor
Solution:
The surface of a perfect conductor is an equipotential surface, which means that the potential difference between any two points on the surface is zero. Therefore, the potential difference across the conductor and water interface is also zero.
Let σ be the surface charge density on the conductor. Then, the electric field just outside the conductor due to this charge is given by:
E = σ/ε0
The electric field just inside the water is given by:
E' = E/ε
Where, ε is the dielectric constant of water.
Since the potential difference across the interface is zero, we have:
E + E' = 0
Or, σ/ε0 + (σ/ε0)/ε = 0
Or, σ/ε0 + σ/(ε0ε) = 0
Or, σ = -εσ/ε0
Or, σ = -ε/ε0 x σ
Substituting the values, we get:
σ = -80 x 8.85 x 10^-12/10^-9 x 36π x 2
σ = 1.14 x 10^-9 C/m^2
Therefore, the surface charge density on the conductor is 1.14 x 10^-9 C/m^2.
Hence, option D is the correct answer.
Electric field on the surface of a perfect conductor = 2 V/m
Dielectric constant of water (ε) = 80ε0
Where, ε0 = 8.85 x 10^-12 F/m
To find: Surface charge density on the conductor
Solution:
The surface of a perfect conductor is an equipotential surface, which means that the potential difference between any two points on the surface is zero. Therefore, the potential difference across the conductor and water interface is also zero.
Let σ be the surface charge density on the conductor. Then, the electric field just outside the conductor due to this charge is given by:
E = σ/ε0
The electric field just inside the water is given by:
E' = E/ε
Where, ε is the dielectric constant of water.
Since the potential difference across the interface is zero, we have:
E + E' = 0
Or, σ/ε0 + (σ/ε0)/ε = 0
Or, σ/ε0 + σ/(ε0ε) = 0
Or, σ = -εσ/ε0
Or, σ = -ε/ε0 x σ
Substituting the values, we get:
σ = -80 x 8.85 x 10^-12/10^-9 x 36π x 2
σ = 1.14 x 10^-9 C/m^2
Therefore, the surface charge density on the conductor is 1.14 x 10^-9 C/m^2.
Hence, option D is the correct answer.
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The electric field on the surface of a perfect conductor is 2V/m. The conductor is immersed in water with ε = 80 ε0. The surface charge density on the conductor is ((ε0 = 10-9/36p)F/m)a)0 c/m2b)2 c/m2c)1.8 x 10-11 c/m2d)1.14 x 10-9 c/m2Correct answer is option 'D'. Can you explain this answer?
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The electric field on the surface of a perfect conductor is 2V/m. The conductor is immersed in water with ε = 80 ε0. The surface charge density on the conductor is ((ε0 = 10-9/36p)F/m)a)0 c/m2b)2 c/m2c)1.8 x 10-11 c/m2d)1.14 x 10-9 c/m2Correct answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The electric field on the surface of a perfect conductor is 2V/m. The conductor is immersed in water with ε = 80 ε0. The surface charge density on the conductor is ((ε0 = 10-9/36p)F/m)a)0 c/m2b)2 c/m2c)1.8 x 10-11 c/m2d)1.14 x 10-9 c/m2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The electric field on the surface of a perfect conductor is 2V/m. The conductor is immersed in water with ε = 80 ε0. The surface charge density on the conductor is ((ε0 = 10-9/36p)F/m)a)0 c/m2b)2 c/m2c)1.8 x 10-11 c/m2d)1.14 x 10-9 c/m2Correct answer is option 'D'. Can you explain this answer?.
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