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An electrostatic charge distribution has the charge density given by rho = Q\{delta(x - x_{0}) - delta(x + x_{0})\} . For this charge distribution, the electric field at (2x_{0}, 0, 0) is?
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An electrostatic charge distribution has the charge density given by r...
Charge Distribution and Electric Field

Charge Distribution:
The charge density of the given electrostatic charge distribution is given by:
ρ = Q[δ(x - x₀) - δ(x + x₀)]

Here, ρ represents charge density, Q represents total charge, δ(x - x₀) and δ(x + x₀) represent Dirac delta functions, and x₀ represents a constant.

The charge density is zero everywhere except at x = x₀ and x = -x₀. At these points, the charge density is Q.

Electric Field:
To find the electric field at a point (2x₀, 0, 0), we can use the principle of superposition. The electric field at this point due to each charge distribution can be calculated separately and then added together.

Electric Field due to δ(x - x₀):
The electric field due to a point charge is given by Coulomb's Law:
E₁ = k * (Q / r₁²)

Here, E₁ represents the electric field due to δ(x - x₀), k represents the electrostatic constant, Q represents the charge, and r₁ represents the distance from the charge to the point (2x₀, 0, 0).

Since the charge distribution is located at x = x₀, the distance from the charge to the point (2x₀, 0, 0) is r₁ = x - 2x₀. Therefore, the electric field due to δ(x - x₀) is:
E₁ = k * (Q / (x - 2x₀)²)

Electric Field due to δ(x + x₀):
Similarly, the electric field due to δ(x + x₀) can be calculated as:
E₂ = k * (Q / (x + 2x₀)²)

Superposition of Electric Fields:
The total electric field at the point (2x₀, 0, 0) is the sum of the electric fields due to δ(x - x₀) and δ(x + x₀):
E_total = E₁ + E₂

Explanation:
The given charge distribution consists of two point charges located at x = x₀ and x = -x₀, with charge density Q. To find the electric field at the point (2x₀, 0, 0), we use the principle of superposition. We calculate the electric field due to each charge separately using Coulomb's Law and then add them together. The electric field due to δ(x - x₀) is given by k * (Q / (x - 2x₀)²), and the electric field due to δ(x + x₀) is given by k * (Q / (x + 2x₀)²). Finally, we add these two electric fields together to obtain the total electric field at the point (2x₀, 0, 0).
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An electrostatic charge distribution has the charge density given by rho = Q\{delta(x - x_{0}) - delta(x + x_{0})\} . For this charge distribution, the electric field at (2x_{0}, 0, 0) is?
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