If you were to separate all of the electrons and protons in 1 g (0.001 kg) of matter, you’d have about 96,000 c of positive charge and the same amount of negative charge. if you placed these charges 4.5 m apart, how strong would the attractive force between them be?

Explanation:

To solve this problem we need to apply the concept relate to Coulomb's law.

We know that the Force given by Coulomb is

Where ,

F= electrostatic force

q= electric charge

r= distance between charge centers

k= Coulomb constant =

We have the values only to replace in the equation, then we have,

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).