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The potential field V = 80r2 cosθ V. The volume charge density at point P(2.5, θ = 300 , Ø = 600 ) in free space is
- a)-2.45 nC/m3
- b)1.42 nC/m3
- c)-1.42 nC/m3
- d)2.45 nC/m3
Correct answer is option 'A'. Can you explain this answer?
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The potential field V = 80r2 cosθ V. The volume charge density at...
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Calculation of Electric Field
Given, potential field V = 80r^2 cosθ
To find the electric field, we need to take the negative gradient of the potential field.
∴ E = -∇V
Taking the gradient, we get
∇V = (dV/dr)er + (1/r)(dV/dθ)eθ
= 160r cosθ er - 80r^2 sinθ eθ
∴ E = -∇V = -160r cosθ er + 80r^2 sinθ eθ
Calculation of Charge Density
Using Gauss's law, we know that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by ε0.
Φ = Q/ε0
Where Φ is the electric flux, Q is the charge enclosed and ε0 is the permittivity of free space.
Assuming a spherical Gaussian surface of radius r with point P at its center, the electric flux passing through the surface is
Φ = ∫E.dS = E(4πr^2)
where E is the electric field at point P.
∴ Q = Φε0 = E(4πr^2)ε0
The volume charge density at point P is given by
ρ = Q/V
where V is the volume enclosed by the Gaussian surface.
For a sphere of radius r, V = (4/3)πr^3
∴ ρ = Q/V = E(4πr^2)ε0 / (4/3)πr^3
Simplifying, we get
ρ = 3Eε0/r
Substituting the values of E and r at point P, we get
ρ = 3(160 × 2.5 × 10^-2 × cos600) × 8.85 × 10^-12 / 2.5
ρ = -2.45 × 10^-9 C/m^3
Note that the negative sign indicates that the charge density is negative, which means that there is an excess of negative charge at point P.
Hence, the correct answer is option A, -2.45 nC/m^3.
Given, potential field V = 80r^2 cosθ
To find the electric field, we need to take the negative gradient of the potential field.
∴ E = -∇V
Taking the gradient, we get
∇V = (dV/dr)er + (1/r)(dV/dθ)eθ
= 160r cosθ er - 80r^2 sinθ eθ
∴ E = -∇V = -160r cosθ er + 80r^2 sinθ eθ
Calculation of Charge Density
Using Gauss's law, we know that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by ε0.
Φ = Q/ε0
Where Φ is the electric flux, Q is the charge enclosed and ε0 is the permittivity of free space.
Assuming a spherical Gaussian surface of radius r with point P at its center, the electric flux passing through the surface is
Φ = ∫E.dS = E(4πr^2)
where E is the electric field at point P.
∴ Q = Φε0 = E(4πr^2)ε0
The volume charge density at point P is given by
ρ = Q/V
where V is the volume enclosed by the Gaussian surface.
For a sphere of radius r, V = (4/3)πr^3
∴ ρ = Q/V = E(4πr^2)ε0 / (4/3)πr^3
Simplifying, we get
ρ = 3Eε0/r
Substituting the values of E and r at point P, we get
ρ = 3(160 × 2.5 × 10^-2 × cos600) × 8.85 × 10^-12 / 2.5
ρ = -2.45 × 10^-9 C/m^3
Note that the negative sign indicates that the charge density is negative, which means that there is an excess of negative charge at point P.
Hence, the correct answer is option A, -2.45 nC/m^3.
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The potential field V = 80r2 cosθ V. The volume charge density at point P(2.5, θ= 300 , Ø= 600 ) in free space isa)-2.45 nC/m3b)1.42 nC/m3c)-1.42 nC/m3d)2.45 nC/m3Correct answer is option 'A'. Can you explain this answer?
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The potential field V = 80r2 cosθ V. The volume charge density at point P(2.5, θ= 300 , Ø= 600 ) in free space isa)-2.45 nC/m3b)1.42 nC/m3c)-1.42 nC/m3d)2.45 nC/m3Correct answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 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 The potential field V = 80r2 cosθ V. The volume charge density at point P(2.5, θ= 300 , Ø= 600 ) in free space isa)-2.45 nC/m3b)1.42 nC/m3c)-1.42 nC/m3d)2.45 nC/m3Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The potential field V = 80r2 cosθ V. The volume charge density at point P(2.5, θ= 300 , Ø= 600 ) in free space isa)-2.45 nC/m3b)1.42 nC/m3c)-1.42 nC/m3d)2.45 nC/m3Correct answer is option 'A'. Can you explain this answer?.
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