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An infinitely long uniform charge of density 30 nC/m is located at y = 3, z = 4. The field intensity at (0, 5, 1) is E. Now, what is the field intensity at (2, 5, 1)?
- a)2E
- b)E/√ 2
- c)E
- d)4E
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
An infinitely long uniform charge of density 30 nC/m is located at y =...
To find the electric field intensity at (2, 5, 1), we can use the principle of superposition.
First, we consider the contribution of the infinitely long uniform charge at y = 3, z = 4 to the electric field at (2, 5, 1).
The distance between the charge and the point (2, 5, 1) is given by:
r = √((2-0)^2 + (5-3)^2 + (1-4)^2) = √(4 + 4 + 9) = √17
The electric field intensity due to an infinitely long uniform charge at a distance r is given by:
E = (k * λ) / r
where k is Coulomb's constant (k = 9 * 10^9 Nm^2/C^2) and λ is the charge density in C/m. In this case, λ = 30 * 10^(-9) C/m.
Substituting the values into the formula, we get:
E = (9 * 10^9 Nm^2/C^2 * 30 * 10^(-9) C/m) / √17
E = 270 / √17 N/C
So the electric field intensity at (2, 5, 1) due to the charge at y = 3, z = 4 is 270 / √17 N/C.
Next, we need to consider the contribution of the infinitely long uniform charge at y = 3, z = 4 to the electric field at (2, 5, 1).
The distance between the charge and the point (2, 5, 1) is the same as before: √17.
Therefore, the electric field intensity at (2, 5, 1) due to the charge at y = 3, z = 4 is also 270 / √17 N/C.
Since the two contributions to the electric field intensity at (2, 5, 1) have the same magnitude and direction, we can simply add them together.
Thus, the total electric field intensity at (2, 5, 1) is 270 / √17 N/C + 270 / √17 N/C = 2 * (270 / √17) N/C.
Therefore, the field intensity at (2, 5, 1) is 2E.
Therefore, the correct answer is a) 2E.
First, we consider the contribution of the infinitely long uniform charge at y = 3, z = 4 to the electric field at (2, 5, 1).
The distance between the charge and the point (2, 5, 1) is given by:
r = √((2-0)^2 + (5-3)^2 + (1-4)^2) = √(4 + 4 + 9) = √17
The electric field intensity due to an infinitely long uniform charge at a distance r is given by:
E = (k * λ) / r
where k is Coulomb's constant (k = 9 * 10^9 Nm^2/C^2) and λ is the charge density in C/m. In this case, λ = 30 * 10^(-9) C/m.
Substituting the values into the formula, we get:
E = (9 * 10^9 Nm^2/C^2 * 30 * 10^(-9) C/m) / √17
E = 270 / √17 N/C
So the electric field intensity at (2, 5, 1) due to the charge at y = 3, z = 4 is 270 / √17 N/C.
Next, we need to consider the contribution of the infinitely long uniform charge at y = 3, z = 4 to the electric field at (2, 5, 1).
The distance between the charge and the point (2, 5, 1) is the same as before: √17.
Therefore, the electric field intensity at (2, 5, 1) due to the charge at y = 3, z = 4 is also 270 / √17 N/C.
Since the two contributions to the electric field intensity at (2, 5, 1) have the same magnitude and direction, we can simply add them together.
Thus, the total electric field intensity at (2, 5, 1) is 270 / √17 N/C + 270 / √17 N/C = 2 * (270 / √17) N/C.
Therefore, the field intensity at (2, 5, 1) is 2E.
Therefore, the correct answer is a) 2E.
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Community Answer
An infinitely long uniform charge of density 30 nC/m is located at y =...
Concept:
The magnitude of electric field intensity at a point P due to infinitely long uniform charge density λ is
r → the perpendicular distance from the point to the infinitely long charge density
Analysis:
Here, the point (0, 5, 1) and (2, 5, 1)
r is the same and is equal to
∴ the electric field will be same in both the cases.
The magnitude of electric field intensity at a point P due to infinitely long uniform charge density λ is
r → the perpendicular distance from the point to the infinitely long charge density
Analysis:
Here, the point (0, 5, 1) and (2, 5, 1)
r is the same and is equal to
∴ the electric field will be same in both the cases.
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