Question

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

Answer 1

2.56 nC

Step-by-step 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:

`C =(Q)/(V) (1)`

• 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:

`Q = \sigma * A (2)`

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

`V = E*d (3)`

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:

`E =(\sigma)/(\epsilon_(0)) (4)`

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

`C= (\epsilon_(0)*A)/(d) (5)`

• 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:

`(Q)/(V) = (\epsilon_(0)*A)/(d) (6)`

• Solving for Q, we get:

`Q = (\epsilon_(0)*A*V)/(d) = (8.85e-12F/m*(0.088m)^(2)*187 V)/(5.0e-3m) = 2.56 nC`