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Let a uniform surface charge density of 5 nC/m2 be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4 and a point charge of 2 μC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) is
- a)10.46 kV
- b)1.98 kV
- c)0.96 kV
- d)3.78 kV
Correct answer is option 'B'. Can you explain this answer?
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Let a uniform surface charge density of 5 nC/m2be present at the z = 0...
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Let a uniform surface charge density of 5 nC/m2be present at the z = 0...
Solution:
Given:
Surface charge density σ = 5 nC/m²
Line charge density λ = 8 nC/m
Point charge q = 2 C
Point A(0, 0, 5)
Point P(2, 0, 0)
Point B(1, 2, 3)
To find:
Potential difference between points A and B i.e. VAB
Concept:
We will use the principle of superposition to find the potential at point B due to each of the charges separately and then add them up to get the total potential.
Calculation:
1. Potential due to the surface charge density σ at point B:
Consider a small element of area dA on the surface charge.
The distance between the element and point B is r = [(1-0)² + (2-0)² + (3-0)²]½ = 3.74 m
The potential due to this element is dV = kdAσ/r
The total potential due to the surface charge is Vσ = ∫dV = kσ/r ∫dA
The surface is infinite in extent, but the potential contribution from any point on the surface will be the same at point B. So, we can assume a circular surface of radius R around point B and then take the limit R → ∞.
Vσ = kσ/r ∫dA = kσ/r ∫2πR²sinθdθdφ = 2πkσR/r
As R → ∞, Vσ → 0 (because σ is finite)
2. Potential due to the line charge density λ at point B:
Consider a small element of length dl on the line charge.
The distance between the element and point B is r = [(1-0)² + (2-0)² + (3-4)²]½ = 2.45 m
The potential due to this element is dV = kλdl/r
The total potential due to the line charge is Vλ = ∫dV = kλ/r ∫dl
The line is infinite in extent, but the potential contribution from any point on the line will be the same at point B. So, we can assume a cylindrical surface of radius R around the line and then take the limit R → ∞.
Vλ = kλ/r ∫dl = 2πkλln(R/r)
As R → ∞, Vλ → ∞ (because λ is infinite)
3. Potential due to the point charge q at point B:
The distance between the point charge and point B is r = [(1-2)² + (2-0)² + (3-0)²]½ = 2.45 m
The potential due to this point charge is Vq = kq/r
4. Total potential at point B:
VAB = VB - VA = Vσ + Vλ + Vq - 0 (because VA = 0)
VAB = 2πkσR/r + 2πkλln(R/r) + kq/r
To find the value of R that minimizes VAB, we differentiate VAB with respect to R and equate it to zero:
dVAB/dR = 2πkσ/r - 2πkλ/r + kq/R² =
Given:
Surface charge density σ = 5 nC/m²
Line charge density λ = 8 nC/m
Point charge q = 2 C
Point A(0, 0, 5)
Point P(2, 0, 0)
Point B(1, 2, 3)
To find:
Potential difference between points A and B i.e. VAB
Concept:
We will use the principle of superposition to find the potential at point B due to each of the charges separately and then add them up to get the total potential.
Calculation:
1. Potential due to the surface charge density σ at point B:
Consider a small element of area dA on the surface charge.
The distance between the element and point B is r = [(1-0)² + (2-0)² + (3-0)²]½ = 3.74 m
The potential due to this element is dV = kdAσ/r
The total potential due to the surface charge is Vσ = ∫dV = kσ/r ∫dA
The surface is infinite in extent, but the potential contribution from any point on the surface will be the same at point B. So, we can assume a circular surface of radius R around point B and then take the limit R → ∞.
Vσ = kσ/r ∫dA = kσ/r ∫2πR²sinθdθdφ = 2πkσR/r
As R → ∞, Vσ → 0 (because σ is finite)
2. Potential due to the line charge density λ at point B:
Consider a small element of length dl on the line charge.
The distance between the element and point B is r = [(1-0)² + (2-0)² + (3-4)²]½ = 2.45 m
The potential due to this element is dV = kλdl/r
The total potential due to the line charge is Vλ = ∫dV = kλ/r ∫dl
The line is infinite in extent, but the potential contribution from any point on the line will be the same at point B. So, we can assume a cylindrical surface of radius R around the line and then take the limit R → ∞.
Vλ = kλ/r ∫dl = 2πkλln(R/r)
As R → ∞, Vλ → ∞ (because λ is infinite)
3. Potential due to the point charge q at point B:
The distance between the point charge and point B is r = [(1-2)² + (2-0)² + (3-0)²]½ = 2.45 m
The potential due to this point charge is Vq = kq/r
4. Total potential at point B:
VAB = VB - VA = Vσ + Vλ + Vq - 0 (because VA = 0)
VAB = 2πkσR/r + 2πkλln(R/r) + kq/r
To find the value of R that minimizes VAB, we differentiate VAB with respect to R and equate it to zero:
dVAB/dR = 2πkσ/r - 2πkλ/r + kq/R² =
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Let a uniform surface charge density of 5 nC/m2be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4 and a point charge of 2 μC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) isa)10.46 kVb)1.98 kVc)0.96 kVd)3.78 kVCorrect answer is option 'B'. Can you explain this answer?
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Let a uniform surface charge density of 5 nC/m2be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4 and a point charge of 2 μC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) isa)10.46 kVb)1.98 kVc)0.96 kVd)3.78 kVCorrect answer is option 'B'. 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 Let a uniform surface charge density of 5 nC/m2be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4 and a point charge of 2 μC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) isa)10.46 kVb)1.98 kVc)0.96 kVd)3.78 kVCorrect answer is option 'B'. 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 Let a uniform surface charge density of 5 nC/m2be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4 and a point charge of 2 μC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) isa)10.46 kVb)1.98 kVc)0.96 kVd)3.78 kVCorrect answer is option 'B'. Can you explain this answer?.
Let a uniform surface charge density of 5 nC/m2be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4 and a point charge of 2 μC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) isa)10.46 kVb)1.98 kVc)0.96 kVd)3.78 kVCorrect answer is option 'B'. 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 Let a uniform surface charge density of 5 nC/m2be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4 and a point charge of 2 μC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) isa)10.46 kVb)1.98 kVc)0.96 kVd)3.78 kVCorrect answer is option 'B'. 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 Let a uniform surface charge density of 5 nC/m2be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4 and a point charge of 2 μC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) isa)10.46 kVb)1.98 kVc)0.96 kVd)3.78 kVCorrect answer is option 'B'. Can you explain this answer?.
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ample number of questions to practice Let a uniform surface charge density of 5 nC/m2be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4 and a point charge of 2 μC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) isa)10.46 kVb)1.98 kVc)0.96 kVd)3.78 kVCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
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