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There are three concentric thin spheres of radius a,b,c (in decreasing order). The total surface charge densities on their surfaces are σ, -σ, σ respectively. The magnitude of electric field at a distance r from the center such that a > r > b is,
- a)0
- b)σ(c2 - b2)/∈r2
- c)σ(a2 - b2)/∈r2
- d)σ(a2 + b2)/∈r2
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
There are three concentric thin spheres of radius a,b,c (in decreasing...
Electric field at a distance r (a > r > b) will be due to charges enclosed in r only, & Since, a sphere acts as a point charge for points outside its surface,
∴ E = kQc/r2 + kQb/r2 = kr2
(σ×4πc2 + (−σ)4πb2)
= σε0r2(c2− b2)
Most Upvoted Answer
There are three concentric thin spheres of radius a,b,c (in decreasing...
Electric field at a distance r (a > r > b) will be due to charges enclosed in r only, & Since, a sphere acts as a point charge for points outside its surface,
∴ E = kQc/r2 + kQb/r2 = kr2
(σ×4πc2 + (−σ)4πb2)
= σε0r2(c2− b2)
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Community Answer
There are three concentric thin spheres of radius a,b,c (in decreasing...
Explanation:
Surface charge distribution:
- The surface charge densities on the spheres are σ, -σ, σ respectively.
- The total surface charge on the spheres is given by Q = 4πr^2σ where r is the radius of the sphere.
Electric field at a distance r:
- To find the electric field at a distance r (where a > r > b), we consider the Gaussian surface as a sphere of radius r.
- Since the electric field due to a charged spherical shell is the same as that of a point charge at the center, we can consider the total charge enclosed by the Gaussian surface.
- The total charge enclosed by the Gaussian surface is Q = 4πa^2σ - 4πb^2σ.
- Applying Gauss's law, the electric field at a distance r is given by E = (Q)/(4πεr^2), where ε is the permittivity of the medium.
Calculation:
- Substituting the values of Q and simplifying, we get E = σ(a^2 - b^2)/εr^2.
- Therefore, the magnitude of the electric field at a distance r from the center such that a > r > b is given by option 'b': σ(a^2 - b^2)/εr^2.
Surface charge distribution:
- The surface charge densities on the spheres are σ, -σ, σ respectively.
- The total surface charge on the spheres is given by Q = 4πr^2σ where r is the radius of the sphere.
Electric field at a distance r:
- To find the electric field at a distance r (where a > r > b), we consider the Gaussian surface as a sphere of radius r.
- Since the electric field due to a charged spherical shell is the same as that of a point charge at the center, we can consider the total charge enclosed by the Gaussian surface.
- The total charge enclosed by the Gaussian surface is Q = 4πa^2σ - 4πb^2σ.
- Applying Gauss's law, the electric field at a distance r is given by E = (Q)/(4πεr^2), where ε is the permittivity of the medium.
Calculation:
- Substituting the values of Q and simplifying, we get E = σ(a^2 - b^2)/εr^2.
- Therefore, the magnitude of the electric field at a distance r from the center such that a > r > b is given by option 'b': σ(a^2 - b^2)/εr^2.
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There are three concentric thin spheres of radius a,b,c (in decreasing order). The total surface charge densities on their surfaces areσ, -σ,σ respectively. The magnitude of electric field at a distance r from the center such that a > r > b is,a)0b)σ(c2- b2)/∈r2c)σ(a2- b2)/∈r2d)σ(a2+b2)/∈r2Correct answer is option 'B'. Can you explain this answer?
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