Question

Two large metal plates of area 1.0 m^2 face each other. They are 5.0 cm apart from each other and have equal but opposite charges on their inner surfaces. If the electric field between the plates, ignoring the edge effects is 55 N C^-1,find the charges on the plates.

Answer 1

the field outside a conductor is related to the surface charge density by E = s/e₀

or, s = e₀E

magnitude of total charge 

q = sA = e₀EA 

= 8.85×10⁻¹² × 60 × 1.0

= 5.31×10⁻¹⁰ C

Answer 2

"The charges on the plate is $4.87 \times 10^{-10} C.$

Given:

Area, $A = 1 m^{2}$

Distance between the plates, d = 5 cm

Electric field, $E = 55 \mathrm{NC}^{-1}$

Charge, q = ?

Solution:

We know that the surface charge density of two current carrying parallel metal plates is given by

$\sigma =E\varepsilon _{ 0 }\rightarrow (1)$

Where,

σ = surface charge density

$\varepsilon _{ 0 }\quad =\quad { Permittivity\quad of\quad freespace }\quad =\quad 8.855\quad \times \quad 10^{ -12 }{ C }^{ -1 }{ m }^{ -2 }$

The surface charge of the plate is given by

$q=\sigma A \rightarrow(2)$

Where,

q = surface charge

Thus, substituting equation 1 in equation 2, we get

$q=E \varepsilon_{0} A$

Substituting the given values, we get,

$q=55 \times 8.855 \times 10^{-12} \times 1$

$q=4.87 \times 10^{-10} C$"