Question

Two 3.09 cm by 3.09 cm plates that form a parallel-plate capacitor are charged to +/- 0.617 nC. What is the electric field strength inside the capacitor if the spacing between the plates is 1.784 mm?

Answer 1

Given:

The charge on the capacitor is Q = 0.617 nC

The distance between plates is d =1.784 mm

The area is

`\begin{gathered} A=\text{ 3.09}*3.09 \\ =9.5481\text{ cm}^2 \end{gathered}`

To find the electric field strength.

Step-by-step explanation:

The electric field strength can be calculated by the formula

`E=(Q)/(\epsilon_oA)`

The constant is

`\epsilon_o=8.85*10^(-12)\text{ C}^2\text{ / N m}^2`

On substituting the values, the electric field strength will be