Answer:
Explanation:
Given
--- Ken's share
Required
The fraction each got
Since they both shared a cake, we have:
Substitute:
Factorize
Divide both sides by 3
Recall that:
What are graphs used for?
1 answer:
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Answer:
* E_ring =
*E_ disk= 2kQ
Explanation:
Let's start by finding the electric field of the charged ring
in the attachment we can see a diagram of the system. Due to circular symmetry, the electric field perpendicular to the axis is canceled and only the electric field remains parallel to the axis.
Eₓ = E cos θ (1)
E = k ∫
cos θ = x / r
using the Pythagorean theorem
r =
we substitute
Eₓ = k ∫
Eₓ = ∫ dq
Eₓ = k \frac{x}{(c^2+y^2)^{3/2} } Q
the ring's electric field
E_ring =
Now let's find the electric field of the disk
The charge is distributed over the entire disk, so let's use the concept of charge density
σ =
Let's approximate the disk as a group of rings, the width of each ring is dr, the area is
dA = 2πr dr
we substitute
σ =
dq = 2π σ r dr
we substitute in equation 1, where the electrioc field is of each ring
Eₓ =
if we use a change of variable
dv = 2rdr
v = r²
Eₓ =
we integrate
Eₓ = k x π σ
we value in the limits from r = 0 to r = R
Eₓ = k π σ x (-2) [ ]
Eₓ = 2π k σ ( )
σ = Q/πR²
substitute
Eₓ = 2 k Q/R² (1 - \frac{x}{(x^2 + R^2 ) ^{1/2} } )
E_ disk= 2kQ
The two electric fields are
* E_ring =
*E_ disk= 2kQ
we can see that the functional relationship of the two fields is different