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A spherical charged conductor has surface density σ of charge. The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charged unchanged then the electric field on the surface of the new sphere will be
- a)4E
- b)E/4
- c)E
- d)E/2
Correct answer is option 'B'. Can you explain this answer?
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A spherical charged conductor has surface densityσof charge. The...
The correct answer is: E/4
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A spherical charged conductor has surface densityσof charge. The...
E = KQ/R2
= Kσ(4πR2)/R2
= σ/ε0
R ----> 2R and σ = unchanged
E' = Kσ[4π(2R)2]/(2R)2
=> σ/ε0 = E
= Kσ(4πR2)/R2
= σ/ε0
R ----> 2R and σ = unchanged
E' = Kσ[4π(2R)2]/(2R)2
=> σ/ε0 = E
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A spherical charged conductor has surface densityσof charge. The...
Understanding Electric Field on a Charged Sphere
When dealing with a spherical charged conductor, the relationship between surface charge density, electric field, and radius is crucial for understanding how changes affect the overall electric field.
Surface Charge Density
- The surface charge density (σ) is defined as the charge per unit area on the surface of the sphere.
- It remains constant when the radius of the sphere is doubled.
Electric Field on the Surface
- The electric field (E) just outside the surface of a charged conductor can be expressed using the formula:
E = σ / ε₀, where ε₀ is the permittivity of free space.
- This formula shows that the electric field on the surface is directly proportional to the surface charge density.
Doubling the Radius
- When the radius of the sphere is doubled (2R), the surface area increases since the surface area of a sphere is given by 4πR².
- The new surface area becomes 4π(2R)² = 16πR², which is four times the original surface area.
Charge Conservation
- If the total charge remains the same, the surface charge density on the new sphere (with radius 2R) will decrease because the same charge is spread over a larger area.
- New surface charge density σ' = Q / (4π(2R)²) = Q / (16πR²) = σ / 4.
Conclusion on Electric Field
- Applying the formula for electric field with the new surface charge density:
E' = σ' / ε₀ = (σ / 4) / ε₀ = E / 4.
- Therefore, the electric field on the surface of the new sphere is E/4, confirming that the correct answer is option 'B'.
When dealing with a spherical charged conductor, the relationship between surface charge density, electric field, and radius is crucial for understanding how changes affect the overall electric field.
Surface Charge Density
- The surface charge density (σ) is defined as the charge per unit area on the surface of the sphere.
- It remains constant when the radius of the sphere is doubled.
Electric Field on the Surface
- The electric field (E) just outside the surface of a charged conductor can be expressed using the formula:
E = σ / ε₀, where ε₀ is the permittivity of free space.
- This formula shows that the electric field on the surface is directly proportional to the surface charge density.
Doubling the Radius
- When the radius of the sphere is doubled (2R), the surface area increases since the surface area of a sphere is given by 4πR².
- The new surface area becomes 4π(2R)² = 16πR², which is four times the original surface area.
Charge Conservation
- If the total charge remains the same, the surface charge density on the new sphere (with radius 2R) will decrease because the same charge is spread over a larger area.
- New surface charge density σ' = Q / (4π(2R)²) = Q / (16πR²) = σ / 4.
Conclusion on Electric Field
- Applying the formula for electric field with the new surface charge density:
E' = σ' / ε₀ = (σ / 4) / ε₀ = E / 4.
- Therefore, the electric field on the surface of the new sphere is E/4, confirming that the correct answer is option 'B'.
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