How & why when a negatively charged plate is kept near a positively charged plate it decreases its potential?

Question

How & why when a negatively charged plate is kept near a positively charged plate it decreases its potential?

Answer 1

1)

V_12 = Q / C

V_12 is the voltage between the plates

Q is the total charge on both plates

C is the capcitance.

Q will remain constant

2)

C = epsilon0 * A/d

epsion0 is a constant

A is a constant in your problem

d is the variable

Put 1) and 2) together and you have:

V_12 is proportional to d

As d gets smaller, so does the voltage.

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