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Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell is
- a)inversely proportional to σ
- b)directly proportional to x2
- c)directly proportional to R
- d)inversely proportional to x2
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Consider a thin spherical shell of radius R consisting of uniform surf...
For a thin uniformly charged spherical shell, the field points outside the shell at a distance x from the centre is
If the radius of the sphere is R, Q = σ4πR2
This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre.
If the radius of the sphere is R, Q = σ4πR2
This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre.
Most Upvoted Answer
Consider a thin spherical shell of radius R consisting of uniform surf...
To find the electric field inside the shell, we can use Gauss's Law. Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0).
In this case, we can consider a Gaussian sphere inside the shell. The electric field at every point on the surface of the Gaussian sphere is perpendicular to the surface, so the electric flux through the Gaussian sphere is simply the electric field multiplied by the surface area of the sphere.
Since the electric field is constant on the surface of the Gaussian sphere, we can write the electric flux as E * 4πr^2, where r is the radius of the Gaussian sphere.
The charge enclosed by the Gaussian sphere is the total charge on the shell, which can be calculated by multiplying the surface charge density by the surface area of the shell.
So, the charge enclosed is σ * 4πR^2.
Applying Gauss's Law, we have:
E * 4πr^2 = σ * 4πR^2 / ε0
Simplifying, we find:
E = σR^2 / (ε0r^2)
So, the electric field inside the shell is inversely proportional to the square of the distance from the center of the shell.
In this case, we can consider a Gaussian sphere inside the shell. The electric field at every point on the surface of the Gaussian sphere is perpendicular to the surface, so the electric flux through the Gaussian sphere is simply the electric field multiplied by the surface area of the sphere.
Since the electric field is constant on the surface of the Gaussian sphere, we can write the electric flux as E * 4πr^2, where r is the radius of the Gaussian sphere.
The charge enclosed by the Gaussian sphere is the total charge on the shell, which can be calculated by multiplying the surface charge density by the surface area of the shell.
So, the charge enclosed is σ * 4πR^2.
Applying Gauss's Law, we have:
E * 4πr^2 = σ * 4πR^2 / ε0
Simplifying, we find:
E = σR^2 / (ε0r^2)
So, the electric field inside the shell is inversely proportional to the square of the distance from the center of the shell.
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Community Answer
Consider a thin spherical shell of radius R consisting of uniform surf...
For a thin uniformly charged spherical shell, the field points outside the shell at a distance x from the centre is
If the radius of the sphere is R, Q = σ4πR2
This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre.
If the radius of the sphere is R, Q = σ4πR2
This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre.
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Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell isa)inversely proportional to σb)directly proportional to x2c)directly proportional to Rd)inversely proportional to x2Correct answer is option 'D'. Can you explain this answer?
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Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell isa)inversely proportional to σb)directly proportional to x2c)directly proportional to Rd)inversely proportional to x2Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell isa)inversely proportional to σb)directly proportional to x2c)directly proportional to Rd)inversely proportional to x2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell isa)inversely proportional to σb)directly proportional to x2c)directly proportional to Rd)inversely proportional to x2Correct answer is option 'D'. Can you explain this answer?.
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Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell isa)inversely proportional to σb)directly proportional to x2c)directly proportional to Rd)inversely proportional to x2Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell isa)inversely proportional to σb)directly proportional to x2c)directly proportional to Rd)inversely proportional to x2Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell isa)inversely proportional to σb)directly proportional to x2c)directly proportional to Rd)inversely proportional to x2Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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