Question

Two parallel plates of area 5.68.10-³ m²

have equal and opposite charges of
4.38 10-¹1 C placed on them. What is
the electric field between the plates?
[?] N/C

Answer 1

`\vec E=871.33 \ (N)/(C) }`

Step-by-step explanation:

Given that two parallel plates with an area of
`5.68 * 10^(-3) \ m^2` have equal and opposite charges of
`4.38 * 10^(-11) \ C`. Find the value of the electric field between them.

Using the following equation,

`\boxed{ \vec E=(Q)/(A \epsilon_0); \ \ \epsilon_0=8.85 * 10^(-12)(C^2)/(Nm^2) }`

Plug the known values into the equation.

`\vec E=(4.38 * 10^(-11))/((5.68 * 10^(-3))(8.85 * 10^(-12)))\\\\\therefore \boxed{\boxed{\vec E=871.33 \ (N)/(C) }}`

Thus, the electric field is found.