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The charge density of a field with a position vector as electric flux density is given by
- a)0
- b)1
- c)2
- d)3
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The charge density of a field with a position vector as electric flux ...
Answer: d
Explanation: The Gauss law for electric field states that the divergence of the electric flux density is the charge density. Thus Div(D) = ρ. For D as a position vector, the divergence of the position vector D will be always 3. Thus the charge density is also 3.
Explanation: The Gauss law for electric field states that the divergence of the electric flux density is the charge density. Thus Div(D) = ρ. For D as a position vector, the divergence of the position vector D will be always 3. Thus the charge density is also 3.
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The charge density of a field with a position vector as electric flux ...
Charge Density in Electric Field:
To determine the charge density of a field with a position vector as electric flux density, we can use the formula:
\[ \rho = \nabla \cdot D \]
Explanation:
- Electric Flux Density (D): Electric flux density (D) is a vector quantity that represents the electric flux per unit area at a given point in space. It is related to the electric field intensity (E) by the equation D = εE, where ε is the permittivity of the material.
- Position Vector: The position vector (r) in this context refers to the vector that locates a point in space with respect to an origin.
- Charge Density (ρ): Charge density is a measure of the amount of electric charge per unit volume at a point in space. It is denoted by the symbol ρ.
- Formula: The formula ρ = ∇ · D represents the charge density in the electric field, where ∇ is the divergence operator.
- Correct Answer (Option D): In this case, the correct answer is option 'D', which is 3. This indicates that the charge density of the field with a position vector as electric flux density is 3.
Therefore, the charge density of the field with a position vector as electric flux density is 3.
To determine the charge density of a field with a position vector as electric flux density, we can use the formula:
\[ \rho = \nabla \cdot D \]
Explanation:
- Electric Flux Density (D): Electric flux density (D) is a vector quantity that represents the electric flux per unit area at a given point in space. It is related to the electric field intensity (E) by the equation D = εE, where ε is the permittivity of the material.
- Position Vector: The position vector (r) in this context refers to the vector that locates a point in space with respect to an origin.
- Charge Density (ρ): Charge density is a measure of the amount of electric charge per unit volume at a point in space. It is denoted by the symbol ρ.
- Formula: The formula ρ = ∇ · D represents the charge density in the electric field, where ∇ is the divergence operator.
- Correct Answer (Option D): In this case, the correct answer is option 'D', which is 3. This indicates that the charge density of the field with a position vector as electric flux density is 3.
Therefore, the charge density of the field with a position vector as electric flux density is 3.
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The charge density of a field with a position vector as electric flux density is given bya)0b)1c)2d)3Correct answer is option 'D'. Can you explain this answer?
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