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The electric field at a distance 3R/2 from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2 from the centre of the sphere is
- a)E/2
- b)zero
- c)E
- d)E/4
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The electric field at a distance 3R/2from the centre of a charged cond...
Electric field at a point inside a charged
conducting spherical shell is zero.
conducting spherical shell is zero.
Most Upvoted Answer
The electric field at a distance 3R/2from the centre of a charged cond...
Explanation:
We have to find the electric field at a distance R/2 from the centre of the charged conducting spherical shell. Let us consider a Gaussian surface of radius R/2 inside the shell. By Gauss's law, the electric flux through this surface is given by:
Φ = q/ε0
where q is the charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
We know that electric field inside a conductor is zero. Therefore, the charge enclosed by the Gaussian surface is zero because there is no charge inside the surface. Hence, the electric flux through the surface is also zero.
Now, let us consider another Gaussian surface of radius 3R/2 outside the shell. By Gauss's law, the electric flux through this surface is given by:
Φ = q/ε0
where q is the charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
The electric field at a distance 3R/2 from the centre of the shell is given as E. Therefore, the charge enclosed by the Gaussian surface is q = 4πε0(E)(3R/2)2 = 9ε0ER2.
Hence, the electric flux through the Gaussian surface is Φ = q/ε0 = 9ER2.
Now, we can use the principle of superposition to find the electric field at a distance R/2 from the centre of the shell. The electric field due to the charged spherical shell is given by:
E = Φ/4πε0R2
Substituting the values of Φ and R, we get:
E = 9E/4
Therefore, the electric field at a distance R/2 from the centre of the shell is zero.
We have to find the electric field at a distance R/2 from the centre of the charged conducting spherical shell. Let us consider a Gaussian surface of radius R/2 inside the shell. By Gauss's law, the electric flux through this surface is given by:
Φ = q/ε0
where q is the charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
We know that electric field inside a conductor is zero. Therefore, the charge enclosed by the Gaussian surface is zero because there is no charge inside the surface. Hence, the electric flux through the surface is also zero.
Now, let us consider another Gaussian surface of radius 3R/2 outside the shell. By Gauss's law, the electric flux through this surface is given by:
Φ = q/ε0
where q is the charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
The electric field at a distance 3R/2 from the centre of the shell is given as E. Therefore, the charge enclosed by the Gaussian surface is q = 4πε0(E)(3R/2)2 = 9ε0ER2.
Hence, the electric flux through the Gaussian surface is Φ = q/ε0 = 9ER2.
Now, we can use the principle of superposition to find the electric field at a distance R/2 from the centre of the shell. The electric field due to the charged spherical shell is given by:
E = Φ/4πε0R2
Substituting the values of Φ and R, we get:
E = 9E/4
Therefore, the electric field at a distance R/2 from the centre of the shell is zero.
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Community Answer
The electric field at a distance 3R/2from the centre of a charged cond...
Electric field inside a charged conductor is always zero.
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The electric field at a distance 3R/2from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2from the centre of the sphere isa)E/2b)zeroc)Ed)E/4Correct answer is option 'B'. Can you explain this answer?
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