Answer:
a) V = k 2π σ (√(b² + x²) - √ (a² + x²))
,
b) E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))
Explanation:
a) The expression for the electric potential is
V = k ∫ dq / r
For this case, consider the disk formed by a series of concentric rings of radius r and width dr, the distance of each ring to point P
R = √(x² + r²)
The charge on a ring is
σ = dq / dA
The area of a ring is
A = π r
dA = 2π r dr
So the charge is
dq = σ 2π r dr
We substitute
V = k σ 2pi ∫ r dr / √(r² + x²)
We integrate
V = k 2π σ √(r² + x²)
We evaluate from the lower limit r = a to the upper limit r = b
V = k 2π σ (√(b² + x²) - √ (a² + x²))
b) the electric field and the potential are related
E = - dV / dx
E = - k 2π σ (1/2 2x /√(b² + x²) - ½ 2x /√(a² + x²))
E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))
I believe the answer is C. protoplanetary disc. Let me know if this helps
As a reference, consider the line from the point perpendicular to the mirror.
That direction is called 'normal' to the mirror.
The ray on the right leaves the point traveling 5° to the right of the normal,
and leaves the mirror on a path that's 10° to the right of the normal.
The ray on the left leaves the point traveling 5° to the left of the normal,
and leaves the mirror on a path that's 10° to the left of the normal.
The angle between the two rays after they leave the mirror is 20° .
Frankly, Charlotte, if there were more than 5 points available for this answer,
I'd seriously consider giving you a drawing too.
Light travels at the speed of 186,000 miles per second. If you were to travel around the earth it would be 7.5 times in a second
