Bound Surface Charge MCQ Questions and Answers
1. What is the main cause of bound surface charge on a polarized insulator?
A. Movement of free electrons through the material
B. Slight separation of positive and negative bound charges
C. Creation of new electric charges
D. Ionization of the entire insulator
Correct Answer:
Slight separation of positive and negative bound charges
Explanation:
Bound surface charge results when polarization slightly separates bound positive and negative charges within the insulating material.
2. The bound surface charge density on a polarized insulator is given by which expression?
A. σ_b = P · n̂
B. σ_b = P + n̂
C. σ_b = P / n̂
D. σ_b = P × n̂
Correct Answer:
σ_b = P · n̂
Explanation:
The bound surface charge density is the dot product of the polarization vector P and the outward unit normal n̂ to the surface.
3. In the expression σ_b = P · n̂, what does n̂ represent?
A. The direction of the electric field
B. The outward unit normal to the surface
C. The direction of motion of free charges
D. The direction of magnetic polarization
Correct Answer:
The outward unit normal to the surface
Explanation:
The vector n̂ is the unit vector directed perpendicular and outward from the surface where the bound charge density is being calculated.
4. A uniformly polarized insulating slab has polarization perpendicular to its two flat faces. On which surfaces is bound surface charge produced?
A. Only on the two faces perpendicular to the polarization
B. Only on the side surfaces parallel to the polarization
C. On all surfaces equally
D. On no surface
Correct Answer:
Only on the two faces perpendicular to the polarization
Explanation:
Bound surface charge is σ_b = P · n̂. It is nonzero where the surface normal is parallel or antiparallel to the polarization and zero on surfaces whose normal is perpendicular to it.
5. For a uniformly polarized insulator, what is the bound surface charge density on a surface whose normal is perpendicular to the polarization?
A. P
B. -P
C. Zero
D. P squared
Correct Answer:
Zero
Explanation:
When the polarization is perpendicular to the surface normal, their dot product is zero, so σ_b = P · n̂ = 0.
6. If the polarization vector points outward from a surface of an insulator, what is the sign of the bound surface charge on that surface?
A. Positive
B. Negative
C. Always zero
D. It must alternate between positive and negative
Correct Answer:
Positive
Explanation:
When P points in the same direction as the outward normal n̂, their dot product is positive, giving positive bound surface charge density.
7. If the polarization vector points inward at a surface of an insulator, what is the sign of the bound surface charge there?
A. Positive
B. Negative
C. Always zero
D. It depends only on the size of the insulator
Correct Answer:
Negative
Explanation:
When P is opposite to the outward normal n̂, the dot product P · n̂ is negative, so the bound surface charge density is negative.
8. A polarized spherical insulator has a uniform polarization directed upward. At which points is the bound surface charge density greatest in magnitude?
A. At the equator, where the surface is parallel to the polarization
B. At the top and bottom of the sphere
C. At every point equally
D. Only at the center of the sphere
Correct Answer:
At the top and bottom of the sphere
Explanation:
The magnitude of σ_b = P · n̂ is greatest where the surface normal is parallel or antiparallel to the uniform polarization, namely at the top and bottom.
9. What is the unit of bound surface charge density?
A. C/m
B. C/m²
C. C/m³
D. N/C
Correct Answer:
C/m²
Explanation:
Bound surface charge density represents charge per unit area, so its SI unit is coulomb per square metre, C/m².
10. For a uniformly polarized finite insulator, the total bound charge over its entire surface is generally
A. Always positive
B. Always negative
C. Zero
D. Equal to the free charge on the surface
Correct Answer:
Zero
Explanation:
Uniform polarization produces positive and negative bound surface charges on different parts of the surface, and their total over the closed surface is zero.
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