A parallel-plate capacitor has an area of 4.0 cm^2, and the plates are separated by 1.0 mm. a. What is the capacitance?
b. How much charge does this capacitor store when connected to a 9.0 V battery?

Explanation:

Capacitance of a parallel-plate capacitor

The parallel plate capacitor consists of two identical conducting plates, each having a surface area A, separated by a distance d with no material between the plates.

The capacitance of the capacitor can be calculated as follows:

Where is the permittivity of free space. Its numerical value in SI units is:

a.

The capacitor of the question has an area of A=4 cm^2 and the plates are separated by d=1 mm. Both magnitudes must be converted to SI:

Now calculate the capacitance:

b.

The charge Q stored in any capacitor C when a voltage V is applied is given by the equation:

Q = CV

Entering the known values into this equation gives:

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!