The plates of a capacitor are charged using a battery, and they produce an electric field across the separation distance d between them. The two plates are now to be pushed together to a separation of d/2. The pushing together can be done either with the battery connected or with it disconnected. In which case, with the battery connected or disconnected. Is the electric field magnitude greater with the battery connected with the battery disconnected?

Explanation:

The equation of the capacitance of the capacitor can be represented as:

Also, the electric field between the plates  can be expressed as:

However;

when pushed to a distance d/2, the new capacitance of the capacitor is:

SImilarly, the new electric field between the plates is:

For Battery disconnected:

The electric field between the plates doesn't rely upon the distance between the plates yet relies upon the magnitude of the charge. At the point when the battery is detached, the charge on the capacitor stays as before, so does the electric field.  

Therefore, the magnitude of the electric field when the battery is associated is twice however much the magnitude of the electric field when the battery is separated and disconnected.

Hence, the ratio is :

Hence, the ratio is = 2

answer: 0

explanation: im   14

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).