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:

yup its correct its: c) b b

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.