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vlabodo
3 months ago
10

A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very fa

r from the wire. What is the ratio Ff/Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?
Physics
1 answer:
Softa [3K]3 months ago
7 0

To address this query, we will utilize principles related to electric fields, linear charge density, and electrostatic forces.

The electric field is described as:

E = \frac{\lambda}{2\pi \epsilon_0 r}

In this context:

\lambda= Linear charge density

\epsilon_0 = Permittivity of free space

r = Distance

The expression for linear charge density can be articulated as:

It is defined as

\lambda = \frac{q}{L}

By substituting,

E = \frac{\frac{q}{L}}{2\pi \epsilon_0 r}

E = \frac{q}{2\pi \epsilon_0 rL}

The initial and final electric force can be represented based on charge and electric field through

F_i = E_i q

F_f = E_f q

Upon substituting the electric field value, we obtain:

F_i = (\frac{q}{2\pi \epsilon_0 rL})q = (\frac{q^2}{2\pi \epsilon_0 rL})

If the length is reduced to one-third, then

F_f = (\frac{q}{2\pi \epsilon_0 r(L/3)})q = (\frac{3q^2}{2\pi \epsilon_0 rL})

The electric force ratio becomes

\frac{F_f}{F_i} = \frac{(\frac{3q^2}{2\pi \epsilon_0 rL})}{(\frac{q^2}{2\pi \epsilon_0 rL})}

\frac{F_f}{F_i} = 3

Hence, the determined ratio is 3.

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3 0
2 months ago
An axle passes through a pulley. Each end of the axle has a string that is tied to a support. A third string is looped many time
Keith_Richards [3271]

Answer:

ΔL = MmRgt / (2m + M)

Explanation:

The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.

ΔL = L − L₀

ΔL = Iω − 0

ΔL = ½ MR²ω

To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.

For the block, two forces act: the weight force mg downward and tension force T upward.

For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.

For the sum of forces in the -y direction on the block:

∑F = ma

mg − T = ma

T = mg − ma

For the sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a/R)

T = ½ Ma

Substituting gives:

mg − ma = ½ Ma

2mg − 2ma = Ma

2mg = (2m + M) a

a = 2mg / (2m + M)

The angular acceleration of the pulley is:

αR = 2mg / (2m + M)

α = 2mg / (R (2m + M))

Finally, the angular velocity after time t is:

ω = αt + ω₀

ω = 2mg / (R (2m + M)) t + 0

ω = 2mgt / (R (2m + M))

Substituting into the previous equations gives:

ΔL = ½ MR² × 2mgt / (R (2m + M))

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3 months ago
When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the
Maru [3345]

Answer:

a

The value at a point inside is Zero

b

The electric field is E = 2.7*10^{6} \ N/C

Explanation:

We know from the problem that

The charge magnitude is q = 3.0 \mu C = 3.0 *10^{-6} \ C

The radius of the spherical ball is r = 5.0 \ mm = 0.005 \ m

According to Gauss’s law, the enclosed charge within a conductor is zero which indicates that the electric field within the spherical ball is zero

On the outside, the electric field around the spherical ball is mathematically expressed as

E = \frac{kq}{ a^2}

Here a denotes a point outside the spherical ball with its value of a = 10 \ cm = \frac{10}{100} = 0.1 \ m

and k represents Coulomb's constant, valued at

k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

=> E = \frac{ 3 *10^{-6} * 9*10^9 }{ (0.1)^2}

=> E = 2.7*10^{6} \ N/C

5 0
1 month ago
A moving sidewalk 95 m in length carries passengers at a speed of 0.53 m/s. One passenger has a normal walking speed of 1.24 m/s
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