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

Answer:

$\vec E=871.33 \ \frac{N}{C} }$

Explanation:

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

Using the following equation,

$\boxed{ \vec E=\frac{Q}{A \epsilon_0}; \ \ \epsilon_0=8.85 \times 10^{-12}\frac{C^2}{Nm^2} }$

Plug the known values into the equation.

$\vec E=\frac{4.38 \times 10^{-11}}{(5.68 \times 10^{-3})(8.85 \times 10^{-12})}\\\\\therefore \boxed{\boxed{\vec E=871.33 \ \frac{N}{C} }}$

Thus, the electric field is found.