(a) 25lx
(b) 11.11lx
<u>Explanation:</u>
Illuminance is inversely proportional to the square of the distance.
So,
where, k is a constant
So,
(a)
If I = 100lx and r₂ = 2r Then,
Dividing both the equation we get
When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx
(b)
If I = 100lx and r₂ = 3r Then,
Dividing equation 1 and 3 we get
When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx
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