Electric Mushroom !

Nice Problem, Loved working on it.

From the image I belive it is clear how I have taken elements and what are my variables.

Now I am using a result,that electric field on a point on the axis at a distance $z$ due to a charged disk having radius $a$ and charge $q$ is given by :

$\displaystyle E = \frac { q }{ 2\pi { \varepsilon }_{ 0 }{ a }^{ 2 } } (1-\frac { z }{ \sqrt { { z }^{ 2 }+{ a }^{ 2 } } } )$

Now force on a elemental charged length $dx$ of wire due elemental charge on the disk element is given by :

$\displaystyle {d}^{2}F = \dfrac { \lambda { H }^{ 2 }dxdq }{ 2\pi { \varepsilon }_{ 0 }{ (R(1-y)) }^{ 2 } } (1-\dfrac { x+y }{ \sqrt { { (x+y) }^{ 2 }+{ \left(\dfrac { (1-y)R }{ H } \right) }^{ 2 } } } )$

Now $\displaystyle dq=\pi \rho dy{ \left(\dfrac { R(1-y) }{ H } \right) }^{ 2 }$

Putting this we have :

$\displaystyle {d}^{2}F = \frac { \lambda \rho }{ 2{ \varepsilon }_{ 0 } } (1-\frac { x+y }{ \sqrt { { (x+y) }^{ 2 }+{ \left(\dfrac { (1-y)R }{ H } \right) }^{ 2 } } } )dxdy$

Finally we get the expression of $F$ as :

$\displaystyle F = \dfrac { \lambda \rho }{ 2{ \varepsilon }_{ 0 } } \int _{ 0 }^{ H }{ \int _{ 0 }^{ \infty }{ (1-\dfrac { x+y }{ \sqrt { { (x+y) }^{ 2 }+{ \left(\dfrac { (1-y)R }{ H } \right) }^{ 2 } } } )dxdy } }$

Putting the value of $\rho$ we get :

$\displaystyle F = \dfrac { 3\lambda Q }{ 2\pi { \varepsilon }_{ 0 }{ R }^{ 2 }H } \int _{ 0 }^{ H }{ \int _{ 0 }^{ \infty }{ (1-\dfrac { x+y }{ \sqrt { { (x+y) }^{ 2 }+{ \left(\dfrac { (1-y)R }{ H } \right) }^{ 2 } } } )dxdy } }$

Putting all the values and simplifying we get :

$F = \displaystyle 54\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ \infty }{ (1-\frac { x+y }{ \sqrt { { (x+y) }^{ 2 }+{ (1-y) }^{ 2 } } } )dxdy } }$

$F = \dfrac { 27ln(1+\sqrt{2}) }{ \sqrt { 2 } } = 16.82 N$

Since you asked to round to the nearest integer hence the answer is $17$

Amazing problem, refreshed a lot of my concepts in one go!

Except i did it by using the electric field of a wire instead of the cones,

the electric field paralell to the wire of a semi infinite wire is given by

$\displaystyle{ \frac {k\lambda sin(\theta)}{h}}$

where $\lambda$ is the charge density and h is the distance from the axes, and $\theta$ is the angle shown in figure as $x$

if we are dealing with a ring of radius R concentric with the axis of the wire, then h = R

so the electric force on a ring of radius R is

$E\times 2\pi R \times \lambda_{2}$

$\displaystyle{ \frac {\lambda \lambda_{2} sin(\theta)}{2\epsilon}}$

where $\lambda_{2}$ is the linear charge density of ring

If the ring is actually a small part of a disc whose surface charge density is $\sigma$ , Then since $\lambda_{2}=\sigma dr$ then the force of interaction is given by

$\dfrac {\sigma \lambda[sin(\theta)]dr]}{2\epsilon}$

Now we are ready to deal with the $cone$

Now to make things treatable, let us right $\sin(\theta)$ interms of distance of the disc from the tip of wire,

using the fact that $\displaystyle{sin(\theta)\quad =\quad \frac { R }{ \sqrt { { R }^{ 2 }+{ z }^{ 2 } } } }$

we rewrite the previous expression for a fixed z as

$\displaystyle{\int _{ 0 }^{ R }{ (\frac { \sigma \lambda }{ 2\epsilon } \frac { R }{ \sqrt { { R }^{ 2 }+{ z }^{ 2 } } } ) } dr\quad =\frac { \sigma \lambda }{2\epsilon} (\sqrt { { R }^{ 2 }+{ z }^{ 2 } } -z)\quad }$

Now we are ready,

finally, the surface charge density of the thin disc is related to that of the cone by the formula

$\rho dz=\sigma$

also R which is the radius of the thin disc can be written in terms of z as

$R=tan(\alpha) (H-z)$ where ' $\alpha$ ' is the half angle

so substituting this and putting proper limits, we get the integral (H is the height of cone)

$\displaystyle{\\ \int _{ 0 }^{ H }{ \frac { \lambda \rho }{ 2\varepsilon } } [\sqrt { (H-z)^{ 2 }{ tan(\alpha) }^{ 2 }+{ z }^{ 2 } } -z]\quad }$

Now putting H=1 and $\displaystyle \tan(\alpha)$ =1

and also that $\displaystyle{\rho \quad =\quad 3\frac { Q }{ \pi { R }^{ 2 }H } \quad }$

$\displaystyle\color{grey}{We\quad get\quad the\quad answer\quad as}$ $\boxed{16.794}$

since nearest integer is asked, Final answer is $\displaystyle\boxed{\lceil 16.794 \rceil= {17}}$