quiver3 using 3D spherical coordinates?

Hi, I've read a lot of posts on this site about using quiver3 to plot a function in non-rectangular coordinates and I am still a little bit confused.

Here is my attempt to plot a 3D function in spherical coordinates using quiver3

if true
[r,theta,phi] = meshgrid(0:0.1:100,-pi/2:pi/100:pi/2, 0:pi/100:pi); %defines meshgrid coverage
%r=(X.^2+Y.^2+Z.^2).^0.5 ;
const=1; %mu_0 * M_4 / 4*pi
Br= const*2*cos(theta)./(r.^3); %mag field in rho
Btheta= const*sin(theta)./(r.^3);%mag field in theta
Bphi=0;
%magnetic field function
figure
quiver3(r,theta,phi,Br,Btheta,Bphi,5)% no defines arrow length, other define direction
%defines arrow direction (u,v,w) at point (x,y,z)
hold on
%surf(X,Y,Z) %produces surface
axis equal %defines axis parameters
hold off
  % code
end

This didn't seem to work (I got an error saying V and W need to be the same size). So I tried converting to rectangular before using the quiver function.

 if true
[r,theta,phi] = meshgrid(0:0.1:100,-pi/2:pi/100:pi/2, 0:pi/100:pi); %defines meshgrid coverage
%r=(X.^2+Y.^2+Z.^2).^0.5 ;
const=1; %mu_0 * M_4 / 4*pi
U= const*2*cos(theta)./(r.^3); %mag field in rho
V= const*sin(theta)./(r.^3);%mag field in theta
W=0;
%magnetic field function
r=(X.^2+Y.^2+Z.^2).^0.5 ;
X=r.*sin(phi).*cos(theta);
Y=r.*sin(phi).*sin(theta);
Z=r.*cos(phi); 
figure
quiver3(X,Y,Z,U,V,W,5)% no defines arrow length, other define direction
%defines arrow direction (u,v,w) at point (x,y,z)
hold on
%surf(X,Y,Z) %produces surface
axis equal %defines axis parameters
hold off
    % code
  end

Which doesn't seem to work either. Is it not possible to use spherical coordinate in quiver or am I doing something wrong?

Thanks, John

[John Draper]

After doing some more reading, I think I realise the answer to my question is no. It is better to convert back into Cartesian and then use the quiver3 function.

[Bin Jiang]

It seems that you only specified one element to W, while it should be something like sz=size(U); W= zeros(sz). Perhaps, there is a better way!