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RideAnS [48]
3 years ago
13

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

f equal magnitude. The distances between A and B, B and C, and C and D are equal. (ref: p.553-561)
a. Which charge experiences the greatest net force? Which charge experiences the smallest net force?

b. Find the ratio of the greatest to the smallest net force.

I really only need help with part b. I just don't understand what the question means when it says '"ratio of greatest to smallest net force" and how I am suppose to show that. Help is much appreciated. : )

Physics
1 answer:
Irina18 [472]3 years ago
5 0

Answer:

a. q_C experiences the greatest net force and q_B experiences the smallest net force

b. Ratio of the greatest to the smallest net force= 9

Explanation:

<u>Electrostatic Forces </u>

Two point-charges q1 and q2, separated a distance d, exert on each other an electrostatic force of magnitude

\displaystyle F=K\frac{q_1q_2}{d^2}

If the charges have the same sign, they repel each other, for different signed charges, they attract. That gives us the direction of each force in the space.

Let's assume all the charges of the problem have a magnitude q, and between two consecutive charges, the distance is d. The proposed layout is shown it the image.

a.

The net force on qA is the sum of those exerted by qB, qC, and qD. But note qB and qC repel qA and qD attracts it, so the total force on qA is

F_{TA}=-F_B-F_C+F_D

Computing the individual forces we have

\displaystyle F_B=\frac{K\ q_A\ q_B}{d^2}=K\ \frac{q^2}{d^2}

\displaystyle F_C=\frac{K\ q_A\ q_C}{(2d)^2}=\frac{1}{4}\ \ \frac{K\ q^2}{d^2}

\displaystyle F_D=\frac{K\ q_A\ q_D}{(3d)^2}=\frac{1}{9}\ \ \frac{K\ q^2}{d^2}

The total force on qA is:

\displaystyle F_{TA}=\frac{K\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})

\displaystyle F_{TA}=-\frac{41}{36}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TA}|=\frac{41}{36}\ \frac{K\ q^2}{d^2}

Charge qA repels qB to the right, qC repels qB to the left, and qD attracts qB to the right, thus

\displaystyle F_{TB}=F_A-F_C+F_D

\displaystyle F_{TB}=\frac{K\ q^2}{d^2}-\frac{K\ q^2}{d^2}+\frac{K\ q^2}{(2d)^2}

\displaystyle F_{TB}=\frac{1}{4}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TB}|=\frac{1}{4}\ \frac{K\ q^2}{d^2}

Charges qA and qb repel qC to the right, and qD attracts qC to the right, thus

\displaystyle F_{TC}=F_A+F_B+F_D

\displaystyle F_{TC}=\frac{K\ q^2}{(2d)^2}+\frac{K\ q^2}{d^2}+\frac{K\ q^2}{d^2}

\displaystyle F_{TC}=\frac{9}{4}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TC}|=\frac{9}{4}\ \frac{K\ q^2}{d^2}

Charge qA and qB attract qD to the left, and qC atracts qD to the left, thus

\displaystyle F_{TD}=-F_A-F_B-F_C

\displaystyle F_{TD}=-\frac{K\ q2}{(3d)^2}-\frac{K\ q2}{(2d)^2}-\frac{K\ q2}{d^2}

\displaystyle F_{TD}=-\frac{49}{36}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TD}|=\frac{49}{36}\ \frac{K\ q^2}{d^2}

Comparing the relative values of all the forces

\displaystyle |F_{TC}|>|F_{TD}|>|F_{TA}|>|F_{TB}|

This means that qc experiences the greatest net force and qB experiences the smallest net force

b.

The ratio of the greatest to the smallest forces is

\displaystyle \frac{|F_{TC}|}{|F_{TB}|}=\frac{\frac{9}{4}}{\frac{1}{4}}=9

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Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur
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Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

\mu_{J,T}=(\frac{dT}{dP})_H

or,

\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}

As per question the formula will be:

\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}   .........(1)

where,

\mu_{J,T} = Joule-Thomson coefficient of the gas = 0.13K/atm

T_1 = initial temperature = 19.0^oC=273+19.0=292.0K

T_2 = final temperature = ?

P_1 = initial pressure = 200.0 atm

P_2 = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

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3 years ago
I need HELP Please !!
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A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa
Nataly_w [17]

Answer:

V_p = 267.258 m/s

V_n = 38.375 m/s      

Explanation:

using the law of the conservation of the linear momentum:

P_i = P_f

where P_i is the inicial momemtum and P_f is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

P_i = m(270 m/s)

P_f = mV_P + M_nV_n

where m is the mass of the proton and V_p is the velocity of the proton after the collision, M_n is the mass of the nucleus and V_n is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = mV_p + 14mV_n

then, m is cancelated and we have:

270 = V_p + 14V_n

This is a elastic collision, so the kinetic energy K is conservated. Then:

K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2

and

Kf = \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

then,

\frac{1}{2}m(270)^2 =  \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

here we can cancel the m and get:

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

now, we have two equations and two incognites:

270 = V_p + 14V_n  (eq. 1)

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

in the second equation, we have:

36450 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2  (eq. 2)

from this last equation we solve for V_n as:

V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

and replace in the other equation as:

270 = V_p + 14\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

so,

V_p = -267.258 m/s

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

V_n = \frac{270+267.258}{14}

V_n = 38.375 m/s  

3 0
3 years ago
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