Why is it important to understand the behavior of light when designing and building any type of structure, such as a house, scho
1 answer:
You might be interested in
297.92J of GPE is added.
Answer:
(a) Charge density σ=6.6375×10²nC/m²
(b) Total charge Q=1.47×10²nC
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
σ=6.6375×10²nC/m²
For (b) total charge Q
Q=σA
Vector is perpendicular to x axis or i component.
Hence i component is 0
j component is 63.5