Question

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?

Answer 1

3.6 x 10^6 N/C

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:

`\begin{gathered} E=(q)/(ε_0\cdot A) \\ \\ \text{ We replacing:} \\ \\ E=(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}`

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