Answer to Question #180945 in Physics for Tin

Question

A square metal plate with a length af0.20 m and with a charge of 9.7x 10 C.

Find the electric field of the square metal plate.

Answer 1

The area of the plate is:

$$A =0.2^2 m^2= 0.040m^2$$

By definition, the charge density is:

$$\sigma = \dfrac{q}{A}$$

where $q = 9.7\times 10C = 97C$ is the charge of the plate.

The electric field of the plate (neglecting side effects) is given as follows (see https://farside.ph.utexas.edu/teaching/316/lectures/node27.html):

$$E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{q}{2\varepsilon_0A}$$

where $\varepsilon_0 = 8.9\times 10^{-12}F/m$ is the vacuum permittivity. Substituting the numbers, obtain:

$$E = \dfrac{97}{2\cdot 8.9\times 10^{-12}\cdot 0.04} \approx 1.4\times 10^{14}V/m$$

Answer. $1.4\times 10^{14}V/m$.