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An infinitely long very thin straight wire carries uniform line charge density 8π × 10–2 C/m. The magnitude of electric displacement vector at a point located 20 mm away from the axis of the wire is ______ C/m2.
Correct answer is '2'. Can you explain this answer?
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An infinitely long very thin straight wire carries uniform line charge...
Π μC/m.
a) Determine the electric field at a distance of 4 cm from the wire.
b) Determine the potential difference between two points, one at a distance of 1 cm from the wire and the other at a distance of 5 cm from the wire.
a) To determine the electric field at a distance of 4 cm from the wire, we can use the formula for the electric field due to a charged wire:
E = λ/(2πεr)
where λ is the line charge density, ε is the permittivity of free space, and r is the distance from the wire.
Plugging in the given values, we get:
E = (8π × 10^-6)/(2π × 8.85 × 10^-12 × 0.04)
E = 45.2 N/C
Therefore, the electric field at a distance of 4 cm from the wire is 45.2 N/C.
b) To determine the potential difference between two points, we can use the formula:
ΔV = -∫E·dr
where ΔV is the potential difference, E is the electric field, and the integral is taken along the path between the two points.
For the given situation, we can choose a path that is perpendicular to the wire, so that the electric field is constant and pointing radially outward from the wire. The magnitude of the electric field is the same as in part (a), so we have:
E = 45.2 N/C
The path length is 4 cm for the first segment (between the wire and the point at 1 cm), and 1 cm for the second segment (between the two points). Therefore, we have:
ΔV = -E(0.04) - E(0.01)
ΔV = -2.25 V
Therefore, the potential difference between the two points is 2.25 V (with the potential at the point 1 cm from the wire being higher than the potential at the point 5 cm from the wire).
a) Determine the electric field at a distance of 4 cm from the wire.
b) Determine the potential difference between two points, one at a distance of 1 cm from the wire and the other at a distance of 5 cm from the wire.
a) To determine the electric field at a distance of 4 cm from the wire, we can use the formula for the electric field due to a charged wire:
E = λ/(2πεr)
where λ is the line charge density, ε is the permittivity of free space, and r is the distance from the wire.
Plugging in the given values, we get:
E = (8π × 10^-6)/(2π × 8.85 × 10^-12 × 0.04)
E = 45.2 N/C
Therefore, the electric field at a distance of 4 cm from the wire is 45.2 N/C.
b) To determine the potential difference between two points, we can use the formula:
ΔV = -∫E·dr
where ΔV is the potential difference, E is the electric field, and the integral is taken along the path between the two points.
For the given situation, we can choose a path that is perpendicular to the wire, so that the electric field is constant and pointing radially outward from the wire. The magnitude of the electric field is the same as in part (a), so we have:
E = 45.2 N/C
The path length is 4 cm for the first segment (between the wire and the point at 1 cm), and 1 cm for the second segment (between the two points). Therefore, we have:
ΔV = -E(0.04) - E(0.01)
ΔV = -2.25 V
Therefore, the potential difference between the two points is 2.25 V (with the potential at the point 1 cm from the wire being higher than the potential at the point 5 cm from the wire).
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An infinitely long very thin straight wire carries uniform line charge density 8π × 10–2 C/m. The magnitude of electric displacement vector at a point located 20 mm away from the axis of the wire is ______ C/m2.Correct answer is '2'. Can you explain this answer?
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