So the 3D convolution operator expects our input to be N x C x D x H x W, where N is our batch size, C is our channels, D is our depth, H is our height, and W is our width. Say that our kernel size is 3, that means for any given value in our output, we utilized a 3x3x3 volume in our DxHxW dimensions across all channels.
For example, suppose that A is our input tensor of size 1 x 8 x 5 x 3 x 3 and our kernel size is 3 with 0 padding and out channels is 8. Let f denote our convolution operation. These will be our following operations done:

f(A[:, :, 0:3, 0:3, 0:3])
f(A[:, :, 1:4, 0:3, 0:3])
f(A[:, :, 2:5, 0:3, 0:3])

The main things to notice here is that each operation takes in all of our channels and only a subset of our D x H x W dimensions. When I say “stride over time”, I really mean “stride over the depth dimension”.
Going back to your problem; we probably want to leave height and width as is. But there’s the possibility of swapping your timestep dimension and your num_channels dimension. The difference boils down to, do we want to compute our convolution across all of timestep or all of num_channels? Another way to ask this question is, do we want to stride over the num_channels dimension or the timestep dimension?
Depending on how we shape our input, the convolution operation will act differently. It might not make a big difference, but I think it’s something worth considering.