Answer:
7. 01 * 10^7 N/C
Explanation:
Parameters given:
Distance, y = 2.9 m
Radius, R = 2.9 cm = 0.029m
Charge, Q = 5.0 µC = 5 * 10^(-6) C
Given that:
E = 2πKσ(1- [y/(R² + y²)]) j^
Charge density, σ, is given as:
σ = Q/A = Q/πR²
=> E = 2πKQ/πR² (1- [y/(R² + y²)]) j^
E = 2KQ/R² (1 - [y/(R² + y²)]) j^
E = (2 * 9 * 10^9 * 5 * 10^(-6) / 0.029²) (1 - [2.9/(0.029² + 2.9²)]) j^
E = 107015.46 * 10^3 *(1 - 2.9/8.41) j^
E = 107015.46 * 10^3 * 0.655 j^
E = 7.01 * 10^7 j^ N/C
The magnitude of the electric field is 7.01 * 10^7 N/C. The j^ shows the direction.
Answer:
Explanation:
L 1: front radius 950 mm, rear radius 2700 mm, refractive index 1.528;
We shall use lens maker's formula , that is
1/f = (μ-1) ( 1/R₁ - 1/R₂) , μ is refractive index of the lens , R₁ and R₂ are radius of curvature of front and rear curved surface.
1/f₁ = (1.528-1)( 1/950 + 1/2700)
f₁ = 1331 mm
L2: front radius 535 mm, rear radius 500 mm, refractive index 1.550.
1/f₂ = (1.550-1)( 1/535 + 1/500)
f₂ = 470 mm
largest angular magnification possible
= f₁ /f₂
= 1331 / 470
= 2.83 ( approx )
Length between two lenses
=1331 +470
= 1801 mm
= 1.8 m Ans
'Displacement' is the distance and direction between the starting point and
ending point, regardless of the path followed to get there.
A particle that's executing simple harmonic motion is always in the same place
where it was one time period ago, and where it will be later after another time
period has passed.
So its displacement during exactly one time period is exactly zero.
