# Problem: Two 2.90 cm × 2.90 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC .What is potential difference across the capacitor if the spacing between the plates is 1.50 mm ?

###### FREE Expert Solution

The charge stored in a capacitor:

$\overline{){\mathbf{Q}}{\mathbf{=}}{\mathbf{C}}{\mathbf{V}}}$

The capacitance of a parallel plate capacitor:

$\overline{){\mathbf{C}}{\mathbf{=}}\frac{\mathbf{k}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}}$

For a vacuum and air, dielectric constant, k = 1.

ε0 = 8.854 × 10-12 F/m

A = l2 = (2.90cm)2(1m/100cm)2 = 8.41 × 10-4 m2

###### Problem Details

Two 2.90 cm × 2.90 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC .

What is potential difference across the capacitor if the spacing between the plates is 1.50 mm ?