# How to calculate distance between two wire-pcb connectors with high voltage for keeping connectors isolation?

I'm planning the way of connecting 8 devices terminal wires all together within a 6kV point (V+), there will be 8 more wires connected to GND (V-). Those V+ and V- points will be in the metal case. The key is how to join 8 wires in an only point inside the case.

I thought bringing V+ and V- (that comes from external source to the case by 2 jacks) until a PCB with only wire-PCB connectors. No more proposal for this PCB.

I have not any experience at HV design, but I' ve learnt some tips and tricks. I have read all the literature that I can.

Searching components, I have found there are wire-pcb connectors that supports until 2500V (wurth). If I understand correctly, this breakdown voltage is refered to the voltage difference that could be supported between pins or closer connectors before isolation falls.

My suggestions:

• So if all the connector pins has 6KV --> Vpin1 - Vpin2 = 0V < 2500V will be ok.
• So if some connector pins are 6KV and other one is 0V --> Vpin1 - Vpin2 = 6KV > 2500V will be a problem and probably it will fail.
• But, I know that E = V/distance and this has much to see with dielectric strength.

Taking the suggestions into account, guessing they were true, I will launch my two question according to the dielectric subject:

1.- Could I separate them -GND connector from 6KV one- some distance between them in order that voltage doesn' t break connectors isolation?

2.- If answer is "yes"; how to calculate mathematically the air gap/distance between two connectors that will support 6KV knowing that they can support 2.5KV?