Two 4.587 cm by 4.587 cm plates that form a parallel-plate capacitor are charged to +/- 0.671 nC. What is the electric field strength inside the capacitor if the spacing between the plates is 1.257 mm?

Respuesta :

ANSWER:

3.6 x 10^6 N/C

STEP-BY-STEP EXPLANATION:

Given:

Charge (q) = 0.671 nC = 0.671 x 10^-9 C

Side (s) = 4.587 cm = 4.587 x 10^-3 m

Vacuum permittivity (ε0) = 8.85 x 10^-12 F/m

We can calculate the electric field using the following formula:

[tex]\begin{gathered} E=\frac{q}{ε_0\cdot A} \\ \\ \text{ We replacing:} \\ \\ E=\frac{0.671\cdot10^{-9}}{(8.85\cdot10^{-12})(4.587\cdot10^{-3})(4.587\cdot10^{-3})} \\ \\ E=\:3603477.12=3.6\cdot10^6\text{ N/C} \end{gathered}[/tex]

The electric field is equal to 3.6 x 10^6 N/C