Answer:
electric field E = - k Q (1 /r(r-a)), force F = - k Q qo / r (r-a) and force for r>>a F ≈ - k Q qo / r²
Explanation:
You are asked to find the electric field of a continuous charge distribution, so we must use the equation
E = k ∫dp /r²
Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.
We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density
λ = q / x
As it is constant, we can write it based on differentials
λ = dq / dx
dq = λ dx
We already have all the terms, let's integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a
E = k ∫ λ dx / x²
E = k la (- 1 / x)
Let's get the negative sign from the parentheses
E = - k λ (1 / x)
E = - k λ (1 /(r-a) -1 /r) = - k λ [a / r (r-a)]
Let's change the charge density with the value of the total charge λ = Q / a
E = - k Q/a [a / r (r-a)]
E = - k Q (1 /r(r-a))
b) We calculate the force.
F = E qo
F = - k Q qo / r (r-a)
c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)
F = - k Qqo / r² (1 - a/r)
F ≈ - k Q qo / r²
velocity, but the kinetic energy is higher among the three because its potential and kinetic energy are higher.