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:

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.