To find the electric field at a point on the x-axis due to two point charges, you can use the principle of superposition. The electric field at a point is the vector sum of the electric fields produced by each charge individually. Here's how you can calculate it:
1. Calculate the electric field (E₁) produced by q₁ at the point x = 2.0 m:
The formula for the electric field produced by a point charge is given by:
E = k * |q| / r²
Where:
- E₁ is the electric field produced by q₁.
- k is the electrostatic constant, approximately 8.99 x 10^9 N m²/C².
- |q| is the magnitude of the charge, |q₁| = 4 μC = 4 x 10⁻⁶ C.
- r is the distance from the charge to the point where you want to find the field. In this case, r₁ = 2.0 m.
Calculate E₁:
E₁ = (8.99 x 10^9 N m²/C²) * (4 x 10⁻⁶ C) / (2.0 m)²
2. Calculate the electric field (E₂) produced by q₂ at the point x = 2.0 m:
The direction of E₂ will be negative because it points from positive charge q₂ toward the point on the x-axis. So, you need to include the negative sign in this case.
E₂ = - (k * |q| / r²)
Where:
- E₂ is the electric field produced by q₂.
- k is the same electrostatic constant.
- |q| is the magnitude of the charge, |q₂| = 4 μC = 4 x 10⁻⁶ C.
- r is the distance from q₂ to the point x = 2.0 m, which is 6.0 m (since q₂ is at x = 8.0 m).
Calculate E₂:
E₂ = - (8.99 x 10^9 N m²/C²) * (4 x 10⁻⁶ C) / (6.0 m)²
3. Find the net electric field (E_net) at x = 2.0 m:
The net electric field at this point is the vector sum of E₁ and E₂. Since they have opposite directions, you'll subtract the magnitudes:
E_net = |E₁| - |E₂|
4. Calculate E_net:
E_net = (E₁ - E₂)
Now, calculate E_net to find the electric field at x = 2.0 m.