# Problem: Two 2.90 cm × 2.90 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC. The electric field is 95082 V/m.What is the potential difference across the capacitor if the spacing between the plates is 3.00 mm ?

###### FREE Expert Solution

Electric field:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{-}\mathbf{∆}\mathbf{V}}{\mathbf{d}}}$

d = 3.00 mm(1m/1000mm) = 3.00 × 10-3 m

###### Problem Details

Two 2.90 cm × 2.90 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC. The electric field is 95082 V/m.

What is the potential difference across the capacitor if the spacing between the plates is 3.00 mm ?