Answer:
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:
[tex]C =\frac{Q}{V} (1)[/tex]
- 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:
[tex]Q = \sigma * A (2)[/tex]
- Assuming an uniform electric field E, the potential difference V can be expressed as follows:
[tex]V = E*d (3)[/tex]
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:
[tex]E =\frac{\sigma}{\epsilon_{0}} (4)[/tex]
- Replacing (4) in (3), and (2) in (1), we can express the capacitance C as follows:
[tex]C= \frac{\epsilon_{0}*A}{d} (5)[/tex]
- 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:
[tex]\frac{Q}{V} = \frac{\epsilon_{0}*A}{d} (6)[/tex]
[tex]Q = \frac{\epsilon_{0}*A*V}{d} = \frac{8.85e-12F/m*(0.088m)^{2}*187 V}{5.0e-3m} = 2.56 nC[/tex]
