Answer:
(a): [tex]k\dfrac{Q}{r(a+r)}.[/tex].
(b): 0
(c):[tex]k\dfrac{qQ}{r(a+r)}.[/tex].
(d): Along positive x axis.
Explanation:
Given that the charge is distributed uniformly on the rod, extended from [tex]x=0[/tex] to [tex]x=a[/tex] and having total charge [tex]Q[/tex].
Let [tex]\lambda[/tex] be the linear charge density of the rod, such that,[tex]\lambda = \dfrac Qa.[/tex]
The electric field due to a charge [tex]q[/tex] at a point [tex]r[/tex] distance away is given by
[tex]E=\dfrac{kq}{r^2}[/tex]
where [tex]k[/tex] is the Coulomb's constant whose value is [tex]9\times 10^9] Nm^2/C^2[/tex].
Now consider a small line element of the rod of length [tex]dx[/tex] at distance [tex]r[/tex] from [tex]x =a+r[/tex].
The electric field at point [tex]x = a + r[/tex] due to this element is given by
[tex]dE = \dfrac{k\ dq}{x^2}[/tex]
[tex]dq[/tex] is the charge on the line charge, then,
[tex]\lambda = \dfrac{dq}{dx}\\\Rightarrow dq=\lambda dx[/tex]
Using this value,
[tex]dE = \dfrac{k\lambda\ dx}{x^2}.[/tex]
The electric field due to the whole rod is given by
[tex]E=\int dE\\=\int\limits_{r}^{a+r} \dfrac{k\lambda\ dx}{x^2}\\=k\lambda\int\limits^{a+r}_{r} \dfrac{ dx}{x^2}\\=k\lambda\int\limits^{a+r}_{r} x^{-2}\ dx\\=k\lambda \left (-x^{-1} \right )\limits^r_{a+r}\ dt\\=-k\lambda \left ( \dfrac1{a+r}-\dfrac 1r\right ) \\=-k\lambda \left ( \dfrac{r-(a+r)}{r(a+r)}\right ) \\=k\lambda \left ( \dfrac{a}{r(a+r)}\right )\\=k\dfrac{Q}{a} \left ( \dfrac{a}{r(a+r)}\right )\\=k\dfrac{Q}{r(a+r)}.[/tex]
This electric is along the x axis only.
(a):
The x component of the electric field at this point = [tex]k\dfrac{Q}{r(a+r)}.[/tex].
(b):
The y component of the electric field at this point = 0.
(c):
The electric field and the electric force are related as
[tex]F=qE.[/tex]
The magnitude of the force that this rod exerts on the charge q = [tex]k\dfrac{qQ}{r(a+r)}.[/tex].
(d):
The direction of this force is along the positive x axis.
