# Problem: Two 2.50 cm × 2.50 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC.a) What is potential difference across the capacitor if the spacing between the plates is 1.50 mm?b) What is the electric field strength inside the capacitor if the spacing between the plates is 1.50 mm?

###### FREE Expert Solution

Capacitance:

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

Potential difference:

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

Electric field:

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

a)

$\mathbit{C}\mathbf{=}\frac{\mathbf{\left(}\mathbf{8}\mathbf{.}\mathbf{85}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{\right)}{\mathbf{\left(}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{×}\mathbf{10}\mathbf{-}\mathbf{2}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}$

94% (413 ratings) ###### Problem Details

Two 2.50 cm × 2.50 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC.

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

b) What is the electric field strength inside the capacitor if the spacing between the plates is 1.50 mm?