Thanks you for anwser me so quickly, 

I will put my entire code here:

> epsilon := 8.85*10^(-12);

> theta := .35;

> L := 1.52;

> alpha := (1/6)*Pi;

> R := L*cos(alpha);

> H := L*sin(alpha);

V := `assuming`([proc (z) options operator, arrow; (1/4)*(int(int(theta*R^3*(H-p)^3/(H^3*L*L*sqrt(R^2*(H-p)^2/H^2+(z-p)^2)), p = 0 .. z), o = 0 .. 2*Pi))/(Pi*epsilon) end proc], [.76 > z and z > 0])

solucionsinbase = fsolve(diff(V(z), z) = 0, z)

It's that correct? I change a little my integral because it was wrong, but any way I use the :=arrow opertator to define my function V(z) and assuming 0<z<0.76 

I hope it is ok

Greetings

Thanks you for anwser me so quickly, 

I will put my entire code here:

> epsilon := 8.85*10^(-12);

> theta := .35;

> L := 1.52;

> alpha := (1/6)*Pi;

> R := L*cos(alpha);

> H := L*sin(alpha);

V := `assuming`([proc (z) options operator, arrow; (1/4)*(int(int(theta*R^3*(H-p)^3/(H^3*L*L*sqrt(R^2*(H-p)^2/H^2+(z-p)^2)), p = 0 .. z), o = 0 .. 2*Pi))/(Pi*epsilon) end proc], [.76 > z and z > 0])

solucionsinbase = fsolve(diff(V(z), z) = 0, z)

It's that correct? I change a little my integral because it was wrong, but any way I use the :=arrow opertator to define my function V(z) and assuming 0<z<0.76 

I hope it is ok

Greetings

Thanks you, I'm very gratefull for this, that fix was what i was missing.

The code works now

Thanks you, I'm very gratefull for this, that fix was what i was missing.

The code works now