Negative charge -Q is distributed uniformly around a quarter-circle of radius a that lies in the first quadrant, with the center of the curvature at the origin. Fine the x and y components of the net electric field at the origin.

Answer 1

To find the x and y components of the net electric field at the origin, we can use the following steps:

1. Divide the quarter-circle into small segments (dQ), each having a small charge dQ = λ * dl, where λ is the linear charge density, and dl is the arc length of the small segment.
2. Calculate the electric field (dE) due to each segment at the origin using the formula dE = k * dQ / r^2, where k is Coulomb's constant and r is the distance between the charge element and the origin.
3. Calculate the x and y components of dE. The angle between the radius and the x-axis is θ. So, the x component of dE is dEx = dE * cos(θ), and the y component of dE is dEy = dE * sin(θ).
4. Integrate dEx and dEy over the total charge, Q.

Using the above steps:

dQ = λ * a * dθ (since dl = a * dθ)

The electric field due to the small charge segment is:
dE = k * dQ / a^2

dEx = dE * cos(θ) = (k * λ * a * dθ * cos(θ)) / a^2
dEy = dE * sin(θ) = (k * λ * a * dθ * sin(θ)) / a^2

Integrating these components over the angle, θ from 0 to π/2:

Ex = (k * λ / a) * ∫(cos(θ)dθ) from 0 to π/2
Ey = (k * λ / a) * ∫(sin(θ)dθ) from 0 to π/2

Solving these integrals:
Ex = (k * λ / a) * [sin(θ)] from 0 to π/2
Ex = (k * λ / a) * (sin(π/2) - sin(0))
Ex = (k * λ / a) * (1 - 0)
Ex = k * λ / a

Ey = (k * λ / a) * [-cos(θ)] from 0 to π/2
Ey = (k * λ / a) * (-cos(π/2) + cos(0))
Ey = (k * λ / a) * (0 + 1)
Ey = k * λ / a

Finally, we want the electric field due to the total charge -Q. Since λ = -Q / (π * a), we can substitute λ into the expressions for Ex and Ey:

Ex = k * (-Q / (π * a)) / a
Ex = -kQ / (π * a^2)

Ey = k * (-Q / (π * a)) / a
Ey = -kQ / (π * a^2)

Thus, the x and y components of the net electric field at the origin are:

Ex = -kQ / (π * a^2)
Ey = -kQ / (π * a^2)