Answer:
(a) Charge density σ=6.6375×10²nC/m²
(b) Total charge Q=1.47×10²nC
Step-by-step explanation:
Given Data
A=47.0 cm =0.47 m
Electric field E=75.0 kN/C
To find
(a) Charge density σ
(b)Total Charge Q
Solution
For (a) charge density σ
From Gauss Law we know that
Φ=Q/ε₀.......eq(i)
Where
Φ is electric flux
Q is charge
ε₀ is permittivity of space
And from the definition of flux
Φ = EA
The flux is electric field passing perpendicularly through the surface
Put the this Φ in equation(i)
EA =Q/ε₀
where Q(charge)=σA
EA=(σA)/ε₀
E=σ/ε₀
σ=ε₀E
![=(8.85*10^(-12) )*(75.0*10^(3) )\\=6.6375*10^(-7) C/m^(2)\\=6.63*10^(2)nC/m^(2)](https://img.qammunity.org/2021/formulas/physics/college/i6fr1dit8i1wjvhghinuidieetemd7n7ab.png)
σ=6.6375×10²nC/m²
For (b) total charge Q
Q=σA
![Q=(6.6375*10^(2) nC/m^(2) )(0.47m)^(2)\\ Q=1.47**10^(2)nC](https://img.qammunity.org/2021/formulas/physics/college/yunbwpmvx43td49yqpl6zua7yxr8ofpep7.png)