In solving for the magnitude of the electric field E⃗(z) produced by a sheet charge with charge density σ, use the planar symmetry since the charge distribution doesn't change if you slide it in any direction of xy plane parallel to the sheet. Therefore at each point, the electric field is perpendicular to the sheet and must have the same magnitude at any given distance on either side of the sheet. To take advantage of these symmetry properties, use a Gaussian surface in the shape of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A which will cancel out of the expression for E(z) in the end. The result of applying Gauss's law to this situation then gives an expression for E(z) for both z>0 and z<0. (Figure 3) Express E(z) for z>0 in terms of some or all of the variables/constants σ, z, and ϵ0.

Check the explanation

Explanation:

Let the charge sheet passes through the middle of the cylinder’s length (which is the distance around the end circles), along z-axis, so the cylinder is perpendicular to the surface. Hence the flux through each end will be.....

kindly check the attached image to see the full explanation to the above question.

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).