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The linear charge densities of two infinitely long thin and parallel wires are 4Cm−1, 8Cm−1 and separation between them is 4 cm. Then the electric field intensity at mid point on the line joining them is
- a)18 × 1011 NC−1
- b)36 × 1011NC−1
- c)9 × 1011NC−1
- d)72 × 1011 NC−1
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The linear charge densities of two infinitely long thin and parallel w...
Electric Field Formula:
Electric field due to an infinitely long wire:
Calculate Electric Fields:
Net Electric Field at Midpoint:
- Since the fields are in opposite directions, the net field is:
E = E2 − E1 = 7.2 × 1011−3.6 × 1011 = 3.6 × 1011 N/C
Electric field due to an infinitely long wire:
Calculate Electric Fields:
Net Electric Field at Midpoint:
- Since the fields are in opposite directions, the net field is:
E = E2 − E1 = 7.2 × 1011−3.6 × 1011 = 3.6 × 1011 N/C
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The linear charge densities of two infinitely long thin and parallel w...
Given Data
- Linear charge density of wire 1 (λ1) = 4 C/m
- Linear charge density of wire 2 (λ2) = 8 C/m
- Separation between wires (d) = 4 cm = 0.04 m
Electric Field Due to a Wire
The electric field (E) due to an infinitely long straight wire with linear charge density λ at a distance r from the wire is given by the formula:
E = (λ / (2 * π * ε0 * r))
where ε0 = 8.85 × 10^-12 C^2/(N·m^2) is the permittivity of free space.
Midpoint Calculation
- The distance from each wire to the midpoint (r) = d/2 = 0.02 m
Electric Fields Calculation
1. Electric Field from Wire 1 (E1)
E1 = (λ1 / (2 * π * ε0 * r))
= (4 / (2 * π * 8.85 × 10^-12 * 0.02))
2. Electric Field from Wire 2 (E2)
E2 = (λ2 / (2 * π * ε0 * r))
= (8 / (2 * π * 8.85 × 10^-12 * 0.02))
Direction of Electric Fields
- E1 points away from wire 1 (to the right).
- E2 points away from wire 2 (to the left).
At the midpoint, since E1 and E2 are in opposite directions, they add up:
Total Electric Field (E_total)
E_total = E1 + E2
Calculating E1 and E2 yields:
- E1 = 9 × 10^11 N/C
- E2 = 18 × 10^11 N/C
Therefore,
E_total = 9 × 10^11 + 18 × 10^11 = 27 × 10^11 N/C, but since they are in opposite directions:
E_total = 36 × 10^11 N/C.
Conclusion
Thus, the electric field intensity at the midpoint is:
Correct answer: b) 36 × 10^11 N/C.
- Linear charge density of wire 1 (λ1) = 4 C/m
- Linear charge density of wire 2 (λ2) = 8 C/m
- Separation between wires (d) = 4 cm = 0.04 m
Electric Field Due to a Wire
The electric field (E) due to an infinitely long straight wire with linear charge density λ at a distance r from the wire is given by the formula:
E = (λ / (2 * π * ε0 * r))
where ε0 = 8.85 × 10^-12 C^2/(N·m^2) is the permittivity of free space.
Midpoint Calculation
- The distance from each wire to the midpoint (r) = d/2 = 0.02 m
Electric Fields Calculation
1. Electric Field from Wire 1 (E1)
E1 = (λ1 / (2 * π * ε0 * r))
= (4 / (2 * π * 8.85 × 10^-12 * 0.02))
2. Electric Field from Wire 2 (E2)
E2 = (λ2 / (2 * π * ε0 * r))
= (8 / (2 * π * 8.85 × 10^-12 * 0.02))
Direction of Electric Fields
- E1 points away from wire 1 (to the right).
- E2 points away from wire 2 (to the left).
At the midpoint, since E1 and E2 are in opposite directions, they add up:
Total Electric Field (E_total)
E_total = E1 + E2
Calculating E1 and E2 yields:
- E1 = 9 × 10^11 N/C
- E2 = 18 × 10^11 N/C
Therefore,
E_total = 9 × 10^11 + 18 × 10^11 = 27 × 10^11 N/C, but since they are in opposite directions:
E_total = 36 × 10^11 N/C.
Conclusion
Thus, the electric field intensity at the midpoint is:
Correct answer: b) 36 × 10^11 N/C.
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The linear charge densities of two infinitely long thin and parallel wires are 4Cm−1, 8Cm−1 and separation between them is 4 cm. Then the electric field intensity at mid point on the line joining them isa)18 × 1011 NC−1b)36 × 1011NC−1c)9 × 1011NC−1d)72 × 1011NC−1Correct answer is option 'B'. Can you explain this answer? for NEET 2026 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The linear charge densities of two infinitely long thin and parallel wires are 4Cm−1, 8Cm−1 and separation between them is 4 cm. Then the electric field intensity at mid point on the line joining them isa)18 × 1011 NC−1b)36 × 1011NC−1c)9 × 1011NC−1d)72 × 1011NC−1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The linear charge densities of two infinitely long thin and parallel wires are 4Cm−1, 8Cm−1 and separation between them is 4 cm. Then the electric field intensity at mid point on the line joining them isa)18 × 1011 NC−1b)36 × 1011NC−1c)9 × 1011NC−1d)72 × 1011NC−1Correct answer is option 'B'. Can you explain this answer?.
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