the electric field strength on the second one is 2.67 N/C.
Explanation:
the electric fiel on the first one is:
E1 = k×q/(r^2)
r^2 = k×q/(E1)
= (9×10^9)×(q)/(24.0)
= 375000000q
then the electric field on the second one is:
E2 = k×q/(R^2)
we know that R = 3r
R^2 = 9×r^2
E2 = k×q/(9×r^2)
= k×q/(9×375000000q)
= k/(9×375000000)
= (9×10^9)/(9×375000000)
= 2.67 N/C
Therefore, the electric field strength on the second one is 2.67 N/C.
f = - 20 cm
Explanation:
This exercise asks us for the focal length, which for a lens in air is
1 / f = (n₂-n₁) (1 / R₁ - 1 / R₂)
where n₂ is the refractive index of the material, n₁ is the refractive index of the medium surrounding the lens, R₁ and R₂ are the radii of the two surfaces.
In this exercise the medium that surrounds the lens is air n₁ = 1 and the lens material has an index of refraction n₂ = n = 1.50, let's substitute in the expression
- 1/40 = (n-1) (1 / R₁ -1 / R₂)
(1 / R₁ - 1 / R₂) = - 1/40 (n-1)
let's calculate
(1 / R₁ -1 / R₂) = - 1/40 (1.50 -1)
(1 / R₁ -1 / R₂) = -1/20
Now we change the construction material for one with refractive index
n = 2, keeping the radii,
1 / f = (n-1) (1 / R₁-1 / R₂)
1 / f = (n-1) (-1/20)
let's calculate
1 / f = (2.00-1) (-1/20)
1 / f = -1/20
f = - 20 cm