Find The Force Of Interaction

First, we find the electric field due to a semi-infinite line charge. In reference to the figure , the Y- axis is in the direction of the line charge.Since , by symmetry the forces along X-axis on the hemisphere would cancel. Hence, we will consider electric field only in the Y- axis direction. At a distance $y$ from origin, we consider a small element of length $dy$ , having charge $\lambda dy$ .

$dE = \frac{\lambda dy}{4 \pi \epsilon_{0} (a^2 + (b + y)^2)}$

$dE_{y} = dE \cos \theta = \frac{\lambda (b+y) dy}{4 \pi \epsilon_{0}(a^2 + (b+y)^2)^{3/2}}$

$\Rightarrow E_{y} = \displaystyle \int_{0}^{\infty} \frac{\lambda (b+y) dy}{(a^2 + (b+y)^2)^{3/2}}$

Putting $(b+y)^2 = t$ , we get

$E_{y} = \frac{\lambda}{8 \pi \epsilon_{0}} \displaystyle \int_{b^2}^{\infty} \frac{dt}{(t+a^2)^{3/2}}$

= $\frac{\lambda}{4 \pi \epsilon_{0} \sqrt{a^2 + b^2}}$

= $\frac{\lambda}{4 \pi \epsilon_{0} r}$

Now we cut a hemispherical shell of radius $r$ and thickness $dr$ with center $O$ having charge $dq = \rho \times 2 \pi r^2 dr$ .

Clearly, $dF = dqE_{y} = \rho \times 2 \pi r^2 dr E_{y}$ = $\frac{2 \pi r^2 \rho \lambda}{4 \pi \epsilon_{0} r} = \frac{\rho \lambda}{2 \epsilon_{0}} r dr$

$F = \displaystyle \int dF = \frac{\rho \lambda}{2 \epsilon_{0}} \int_{0}^{R} r dr = \boxed{\frac{\rho \lambda R^2}{4 \epsilon_{0}}}$

Putting values, we get $F = 278.835$

The trick is to first apply newtons 3rd law and realise that it does not matter whose electric field's force on whom u take, it will be the same,, and since it is much easier for the semi infinite wire, than for the complex hemisphere for which we have to first divide into discs then rings,, amazing question :)