Answer :
Answer:
E = k Q 1 / (x₀-x₂) (x₀-x₁)
Explanation:
The electric field is given by
dE = k dq / r²
In this case as we have a continuous load distribution we can use the concept of linear density
λ= Q / x = dq / dx
dq = λ dx
We substitute in the equation
∫ dE = k ∫ λ dx / x²
We integrate
E = k λ (-1 / x)
We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁
E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)
E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)
We replace the density
E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]
E = k Q 1 / (x₀-x₂) (x₀-x₁)
The magnitude of electric field at x0 on the x-axis is E = k Q 1 / (x₀-x₂) (x₀-x₁).
The electric field can be calculated by
dE = k dq / r²
Where,
dE - electric field
k - Coulomb constant of 8.98755 × 109 N · m2 /C 2
dq - Charge
r - distance between two point
The linear density
λ = Q / x = dq / dx
dq = λ dx
Put the values in the equation,
∫ dE = k ∫ λ dx / x²
Now, integrate
E = k λ (-1 / x)
Calculate, between the lower limits x = x₀- x₂ and higher x = x₀-x₁
E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)
E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)
Replace the density
E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]
E = k Q 1 / (x₀-x₂) (x₀-x₁)
Therefore, the magnitude of electric field at x0 on the x-axis is E = k Q [1 / (x₀-x₂) (x₀-x₁)].
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