physics : class12 Superposition in Continuous Charge Distributions : Superposition Principle : Electrostatics MCQ Questions and Answers
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Superposition in Continuous Charge Distributions
MCQ questions and answers.
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MCQ Questions
1. A continuous charge distribution is divided into many small elements dq. According to the superposition principle, the total electric field at a point is found by
A.
Adding the electric fields due to all charge elements vectorially
B.
Adding only the magnitudes of all charge elements
C.
Using the field due to the largest charge element only
D.
Multiplying the field of one element by the total charge
Correct Answer:
Adding the electric fields due to all charge elements vectorially
Explanation:
For a continuous charge distribution, each small element dq produces a small electric field dE. The total field is the vector sum, or integral, of all these contributions.
2. For a continuous line charge with linear charge density λ, a small charge element of length dl is represented by
A.
dq = λ dl
B.
dq = λ/dl
C.
dq = dl/λ
D.
dq = λ + dl
Correct Answer:
dq = λ dl
Explanation:
Linear charge density is charge per unit length, so the charge on a small length dl is dq = λ dl. This element is used when summing the fields of a continuous line charge.
3. The electric field due to a continuous charge distribution is generally calculated using
A.
A sum or integral of the fields due to small charge elements
B.
Only the total charge divided by the distance
C.
The charge density without considering position
D.
The field due to one arbitrarily selected charge element
Correct Answer:
A sum or integral of the fields due to small charge elements
Explanation:
Superposition requires adding the contributions from every small element. In a continuous distribution, this addition becomes an integration over the distribution.
4. For a continuous surface charge distribution with surface charge density σ, the charge on a small area element dA is
A.
dq = σ dA
B.
dq = σ/dA
C.
dq = dA/σ
D.
dq = σ + dA
Correct Answer:
dq = σ dA
Explanation:
Surface charge density is charge per unit area. Therefore, a small surface element dA contains charge dq = σ dA, whose electric field contributes to the total field by superposition.
5. For a continuous volume charge distribution with volume charge density ρ, the charge in a small volume element dV is
A.
dq = ρ dV
B.
dq = ρ/dV
C.
dq = dV/ρ
D.
dq = ρ + dV
Correct Answer:
dq = ρ dV
Explanation:
Volume charge density is charge per unit volume. Thus, the small charge element used in superposition is dq = ρ dV.
6. When calculating the electric field of a continuous charge distribution, the field contribution from a small charge element depends on
A.
The charge of the element and its distance and direction from the point
B.
Only the total charge of the distribution
C.
Only the shape of the observation point
D.
Only the number of charge elements
Correct Answer:
The charge of the element and its distance and direction from the point
Explanation:
Each element produces a field whose magnitude depends on its charge and distance, while its direction is along the line joining the element to the observation point. These vector contributions are then superposed.
7. A uniformly charged rod is treated as a continuous distribution. To find its electric field at a point, the rod is divided into
A.
Small charge elements dq
B.
Only two equal charges
C.
A single point charge at its center
D.
Small magnetic elements
Correct Answer:
Small charge elements dq
Explanation:
A continuous rod is analyzed by considering many small charge elements dq. The electric fields due to these elements are added by integration.
8. If the electric field contributions from different elements of a continuous charge distribution point in the same direction, the total field is obtained by
A.
Adding their magnitudes
B.
Subtracting all their magnitudes
C.
Using only the smallest contribution
D.
Using only the direction of the largest contribution
Correct Answer:
Adding their magnitudes
Explanation:
When all elemental fields are parallel and point in the same direction, their vector sum equals the sum of their magnitudes. This is a direct application of superposition.
9. The electric field of a continuous charge distribution at a point can be written in elemental form as
A.
dE = (1/(4πε₀)) dq/r²
B.
dE = (1/(4πε₀)) r²/dq
C.
dE = (1/(4πε₀)) dq r²
D.
dE = (1/(4πε₀)) r/dq
Correct Answer:
dE = (1/(4πε₀)) dq/r²
Explanation:
Each small charge element dq produces an electric field whose magnitude follows Coulomb's law, dE = (1/(4πε₀)) dq/r². The total field is found by integrating these elemental contributions with their directions.
10. For a continuous charge distribution, replacing the summation of fields of many small charges by an integral is valid because
A.
The charge is spread continuously
B.
The electric field becomes zero
C.
All charge elements have equal distance from every point
D.
The charge density must always be zero
Correct Answer:
The charge is spread continuously
Explanation:
A continuous distribution contains charge throughout a length, area, or volume. Therefore, the discrete superposition sum is replaced by an integral over the small charge elements.
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