A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on the x-axis? Assume that ke = 1 4 π ǫ0 and x0 > x2 > x1 for a Coulomb constant of 8.98755 × 109 N · m2 /C 2

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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  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)

c.the charges if the proton and electron and gravity

explanation:

newton's third law, for every action there is an equal and opposite reaction.