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2 months ago

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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?

1 answer:

Softa [3K]2 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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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

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Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

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Additional explanation

The Ideal Gas Law that should be remembered is:

$\large {\boxed {PV = nRT} }$

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

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Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

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Next, we will determine the work performed on the gas:

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

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Step B:

By utilizing the methodology mentioned earlier:

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

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Next, we will ascertain the work completed on the gas:

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

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Ultimately, we can calculate the total work done and heat supplied as follows:

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

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$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

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Answer details

Grade: High School

Subject: Physics

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Response:

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