What happens to the local surface charge density when the radius of curvature is at its minimum?
Irregularly Shaped Conductors in Electrostatic Equilibrium

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Physics, Science, Engineering, Chemistry
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11th Grade - University
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Hard
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5 questions
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1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
It decreases.
It reaches its maximum.
It becomes zero.
It remains constant.
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
In the example of two conducting spheres, what is the primary reason for charge redistribution?
To equalize the mass of the spheres.
To ensure both spheres have the same electric potential.
To minimize the distance between the spheres.
To maximize the charge on the larger sphere.
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
How quickly does the charge redistribution occur when two conductors touch?
It takes a few hours.
It happens almost instantaneously.
It takes several minutes.
It never completes.
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the relationship between the electric field and the radius of curvature on the surface of a conductor?
The electric field is stronger where the radius of curvature is smaller.
The electric field is weaker where the radius of curvature is smaller.
The electric field is independent of the radius of curvature.
The electric field is the same regardless of the radius of curvature.
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What does the electric field at the surface of a conductor in electrostatic equilibrium equal?
The local surface charge density divided by permittivity of free space.
The local surface charge density minus permittivity of free space.
The local surface charge density multiplied by permittivity of free space.
The local surface charge density plus permittivity of free space.
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