An infinitely long line charge of uniform linear charge density λ = -3.00 µC/m lies parallel to the y axis at x = -2.00 m. A point charge of 4.70 µC is located at x = 1.50 m, y = 2.50 m. Find the electric field at x = 2.50 m, y = 2.00 m.
I get 23.7 kN/C, why is that wrong?
I haven't done the calcs, but how can you just have a magnitude? This is a vector addition of two components of E.
yes, I added the vectors, of the charges. for a uniform line, E= kλ/R. Since the charge on the line is negative, it goes towards the line, and only has an x component of E= -12 kN/C. For the point charge, which is negative, E is away from the source, I got E=30.2 kN/C for the x component, and -15.1 kN/c for the y component. I added the x components, and then found the final magnitude with sqrt(x^2+y^2) as 23.7kN/C. But that is the wrong answer.
To find the electric field at a particular point due to a line charge and a point charge, we can calculate the electric field produced by each charge separately and then perform vector addition to find the total electric field.
The electric field produced by an infinitely long line charge is given by the formula:
E_line = (λ / (2 * π * ε0 * r)) * k
Where:
λ = linear charge density of the line charge (-3.00 µC/m in this case)
r = distance from the line charge to the point where you want to find the electric field (in this case, r = x - x_line_charge, or r = 2.50 m - (-2.00 m) = 4.50 m)
k = Coulomb's constant (9.00 x 10^9 Nm²/C²)
ε0 = permittivity of free space (8.85 x 10^-12 C²/Nm²)
Plugging in the given values, we have:
E_line = ((-3.00 x 10^-6 C/m) / (2 * π * (9.00 x 10^9 Nm²/C²) * (4.50 m))) * k
E_line = (-333.33 N/C) * k
Now, let's calculate the electric field produced by the point charge using the formula:
E_point = (k * Q) / r²
Where:
Q = charge of the point charge (4.70 µC in this case)
r = distance from the point charge to the point where you want to find the electric field (in this case, r = √((x - x_point)² + (y - y_point)²), or r = √((2.50 m - 1.50 m)² + (2.00 m - 2.50 m)²) = 1.12 m)
Plugging in the given values, we have:
E_point = ((9.00 x 10^9 Nm²/C²) * (4.70 x 10^-6 C)) / (1.12 m)² * k
E_point = (375.00 N/C) * k
Now, let's find the total electric field at the point (2.50 m, 2.00 m) by adding the electric fields due to the line charge and the point charge:
E_total = E_line + E_point
E_total = (-333.33 N/C) * k + (375.00 N/C) * k
E_total = 41.67 N/C * k
Hence, the correct value for the electric field at (2.50 m, 2.00 m) is 41.67 N/C. Therefore, 23.7 kN/C is incorrect, and the correct answer is 41.67 N/C.
To find the electric field at a specific point, you need to use the principle of superposition. The total electric field at a point is the vector sum of the electric fields produced by each charge.
In this question, there are two charges: the line charge and the point charge. The line charge produces an electric field, and the point charge produces another electric field.
To find the electric field produced by the line charge, we can use the formula for the electric field of an infinitely long line charge. The electric field at a perpendicular distance r from the line charge is given by:
E_line = λ / (2πε₀r)
where λ is the linear charge density, ε₀ is the permittivity of free space (8.85 x 10^-12 C²/(N·m²)), and r is the distance from the point to the line charge.
In this case, the line charge is located at x = -2.00 m. To find the electric field at x = 2.50 m, y = 2.00 m, we need to calculate the electric field produced by the line charge and the electric field produced by the point charge at that point.
Let's calculate the electric field due to the line charge first:
E_line = λ / (2πε₀r)
λ = -3.00 µC/m
r = distance from the line charge to the point (2.50 m - (-2.00 m)) = 4.50 m
E_line = (-3.00 x 10^-6 C/m) / (2π(8.85 x 10^-12 C²/(N·m²))(4.50 m))
E_line = -2.40 x 10^9 N/C
Now, let's calculate the electric field due to the point charge using the formula:
E_point = k * q / r²
where k is the electrostatic constant (9 x 10^9 N·m²/C²), q is the charge, and r is the distance from the point charge to the point of interest.
q = 4.70 µC
r = distance from the point charge to the point (2.50 m - 1.50 m) = 1.00 m
E_point = (9 x 10^9 N·m²/C²)(4.70 x 10^-6 C) / (1.00 m)²
E_point = 4.70 x 10^4 N/C
To find the total electric field, we need to add the electric fields due to both charges:
E_total = E_line + E_point
E_total = -2.40 x 10^9 N/C + 4.70 x 10^4 N/C
E_total ≈ -2.40 x 10^9 N/C
So the correct answer is approximately -2.40 x 10^9 N/C, not 23.7 kN/C.