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Vesna [10]
4 years ago
9

A uniformly charged ring of radius 10.0 cm has a total charge of 66.0 μC. Find the electric field on the axis of the ring at the

following distances from the center of the ring. (Choose the x-axis to point along the axis of the ring.) (a) 1.00 cm
Physics
1 answer:
Tems11 [23]4 years ago
5 0

To solve this problem we will apply the mathematical consideration of the electric field on an axial axis of a ring. This definition is already established mathematically and is a subordinate of the definition of the magnetic field of Coulomb's laws. It can be expressed as,

E = \frac{kQx}{\sqrt{(x^2+R^2)^3}}

Here,

k = Coulomb's constant

Q = Charge

x = Distance on the axial line

R = Radius of the circle or ring

At x = 1 cm, we have that the total charge is 66.0 μC and the radius 0.1m, then replacing,

E = \frac{(9*10^9)(66*10^{-6})(0.01)}{\sqrt{((0.01)^2+(0.1)^2)^3}}

E = 5.85*10^6 N/C

Therefore the electric field on the axis of the ring is 5.85MN/C

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Answer:

Fraction = 59049/60025

Explanation:

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Now, formula for kinetic energy before collision is;

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Formula for kinetic energy after collision is;

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v is the velocity of the neutron after collision.

From collision principle where momentum before collision equals momentum after collision, we can say that;

(m - M)u = (m + M)v

Thus,

v = [(m - M)u]/(m + M)

Putting [(m - M)u]/(m + M) for v in the final kinetic energy equation gives;

K_f = ½m([(m - M)u]/(m + M))²

K_f = ½mu²((m - M)²/(m + M)²)

To get the fraction of the neutron's kinetic energy is transferred to the plutonium nucleus, it is simply;

K_f/K_i = [½mu²((m - M)²/(m + M)²)]/½mu²

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But mass of plutonium = 244m

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K_f/K_i = ((m - 244m)²/(m + 244m)²)

K_f/K_i = 59049m²/60025m²

K_f/K_i = 59049/60025

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