Answer to Question #180945 in Physics for Tin

Question #180945

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.


1
Expert's answer
2021-04-15T10:32:38-0400

The area of the plate is:


A=0.22m2=0.040m2A =0.2^2 m^2= 0.040m^2

By definition, the charge density is:


σ=qA\sigma = \dfrac{q}{A}

where q=9.7×10C=97Cq = 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=σ2ε0=q2ε0AE = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{q}{2\varepsilon_0A}

where ε0=8.9×1012F/m\varepsilon_0 = 8.9\times 10^{-12}F/m is the vacuum permittivity. Substituting the numbers, obtain:


E=9728.9×10120.041.4×1014V/mE = \dfrac{97}{2\cdot 8.9\times 10^{-12}\cdot 0.04} \approx 1.4\times 10^{14}V/m

Answer. 1.4×1014V/m1.4\times 10^{14}V/m.


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