A parallel plate capacitor is made up of two metal squares with sides of length 8.8 cm, separated by a distance 5.0 mm. When a voltage 187 V is set up across the terminals of the capacitor, the charge stored on the positive plate is equal to nC. g

2.56 nC

Explanation:

By definition, the capacitance is expressed by the following relationship between the charge stored on one of the plates of the capacitor and the potential difference between them, as follows:

       

For a parallel-plate capacitor, assuming a uniform surface charge density σ, if the area of the plates is A, the charge on one of the plates can be written as follows:

       

Assuming an uniform electric field E, the potential difference V can be expressed as follows:

       

        where d is the distance between plates.

Applying Gauss 'Law to a closed surface half within one plate, half outside it, we find that E can be written as follows:

       

Replacing (4) in (3), and (2) in (1), we can express the capacitance C as follows:

       

Taking (1) and (5), as both left sides are equal each other, the right sides are also equal, so we can write the following equality:

       

Solving for Q, we get:

       

Ihope you find the answer