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It is known that the potential is given by V = 70 r0.6 V. The volume charge density at r = 0.6 m is
- a)1.79 nC/m3
- b)-1.79 nC/m3
- c)1.22 nC/m3
- d)-1.22 nC/m3
Correct answer is option 'D'. Can you explain this answer?
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It is known that the potential is given by V = 70 r0.6V. The volume ch...
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It is known that the potential is given by V = 70 r0.6V. The volume ch...
Understanding the Potential Function
The given potential is V = 70 r^0.6 V. To find the volume charge density (ρ) at r = 0.6 m, we can use Poisson's equation, which relates the electric potential to charge density.
Poisson's Equation
Poisson’s equation states:
∇²V = -ρ/ε₀
Where:
- ∇²V is the Laplacian of the potential,
- ρ is the volume charge density,
- ε₀ is the permittivity of free space.
In spherical coordinates, the Laplacian for a spherically symmetric potential is:
∇²V = (1/r²) * d/dr(r² * dV/dr)
Calculating the Derivatives
1. First Derivative (dV/dr):
- Differentiate V = 70 r^0.6.
- dV/dr = 70 * 0.6 * r^(-0.4) = 42 r^(-0.4).
2. Second Derivative (d²V/dr²):
- Differentiate dV/dr again.
- d²V/dr² = -42 * 0.4 * r^(-1.4) = -16.8 r^(-1.4).
3. Substituting into the Laplacian:
- Now substituting into the Laplacian formula leads to:
- ∇²V = (1/r²) * d/dr(r² * dV/dr)
- Substitute r = 0.6 m into the above expression.
Calculating Charge Density
After performing these calculations, we find:
- ρ = -ε₀ * ∇²V.
Substituting ε₀ and the value of ∇²V will lead to the volume charge density at r = 0.6 m.
Final Result
Upon calculation, the volume charge density is found to be -1.22 nC/m³, which corresponds with option 'D'. This negative sign indicates that the charge density is negative, consistent with the nature of the potential function provided.
The given potential is V = 70 r^0.6 V. To find the volume charge density (ρ) at r = 0.6 m, we can use Poisson's equation, which relates the electric potential to charge density.
Poisson's Equation
Poisson’s equation states:
∇²V = -ρ/ε₀
Where:
- ∇²V is the Laplacian of the potential,
- ρ is the volume charge density,
- ε₀ is the permittivity of free space.
In spherical coordinates, the Laplacian for a spherically symmetric potential is:
∇²V = (1/r²) * d/dr(r² * dV/dr)
Calculating the Derivatives
1. First Derivative (dV/dr):
- Differentiate V = 70 r^0.6.
- dV/dr = 70 * 0.6 * r^(-0.4) = 42 r^(-0.4).
2. Second Derivative (d²V/dr²):
- Differentiate dV/dr again.
- d²V/dr² = -42 * 0.4 * r^(-1.4) = -16.8 r^(-1.4).
3. Substituting into the Laplacian:
- Now substituting into the Laplacian formula leads to:
- ∇²V = (1/r²) * d/dr(r² * dV/dr)
- Substitute r = 0.6 m into the above expression.
Calculating Charge Density
After performing these calculations, we find:
- ρ = -ε₀ * ∇²V.
Substituting ε₀ and the value of ∇²V will lead to the volume charge density at r = 0.6 m.
Final Result
Upon calculation, the volume charge density is found to be -1.22 nC/m³, which corresponds with option 'D'. This negative sign indicates that the charge density is negative, consistent with the nature of the potential function provided.
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It is known that the potential is given by V = 70 r0.6V. The volume charge density at r = 0.6 m isa)1.79 nC/m3b)-1.79 nC/m3c)1.22 nC/m3d)-1.22nC/m3Correct answer is option 'D'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2026 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about It is known that the potential is given by V = 70 r0.6V. The volume charge density at r = 0.6 m isa)1.79 nC/m3b)-1.79 nC/m3c)1.22 nC/m3d)-1.22nC/m3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for It is known that the potential is given by V = 70 r0.6V. The volume charge density at r = 0.6 m isa)1.79 nC/m3b)-1.79 nC/m3c)1.22 nC/m3d)-1.22nC/m3Correct answer is option 'D'. Can you explain this answer?.
It is known that the potential is given by V = 70 r0.6V. The volume charge density at r = 0.6 m isa)1.79 nC/m3b)-1.79 nC/m3c)1.22 nC/m3d)-1.22nC/m3Correct answer is option 'D'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2026 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about It is known that the potential is given by V = 70 r0.6V. The volume charge density at r = 0.6 m isa)1.79 nC/m3b)-1.79 nC/m3c)1.22 nC/m3d)-1.22nC/m3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for It is known that the potential is given by V = 70 r0.6V. The volume charge density at r = 0.6 m isa)1.79 nC/m3b)-1.79 nC/m3c)1.22 nC/m3d)-1.22nC/m3Correct answer is option 'D'. Can you explain this answer?.
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