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Shtirlitz [24]
3 years ago
6

three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a

t x=+4.00 cm. charge q1 is located at x+2.00 cm. what is the magnitude of q1 if the net force on q3 is zero
Physics
1 answer:
valentinak56 [21]3 years ago
5 0

Answer:

q_1=+0.375\ {10}^{-9}

Explanation:

Electrostatic Forces

The force exerted between two point charges q_1 and q_2 separated a distance d is given by Coulomb's formula

\displaystyle F=\frac{k\ q_1\ q_2}{d^2}

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

\displaystyle q_3=+5\ 10^{-9}\ c

\displaystyle q_2=-3\ 10^{-9}\ c

The distance between q_3 and q_2 is

\displaystyle d_2=4cm=0.04\ m

The distance between q_3 and q_1 is

\displaystyle d_1=2cm=0.02\ m

We must find the value of q_1 such that

\displaystyle |F_3|=0

Applying Coulomb's formula for q_1 is

\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}

Now for q_2

\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}

If the total force on q_3 is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

\displaystyle F_{13}=F_{23}

\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}

Simplfying and solving for q_1

\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}

\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}

\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}

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a) V_{wf} = 4.67m/s

b) V = 8.29 m/s

Explanation:

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The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

m_{b}V{b_{i} }  + V_{wi}  = m_{w} V_{wf} + m_{b}V_{bf}

where m_{b},V{b_{i} are the mass and the initial velocity of the bullet, m_{w} and V_{wi} are the mass and the initial velocity of the wooden block, and V_{wf} and V_{bf} are the final velocities of the wooden block and the bullet

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m_{b}V{b_{i} }  = m_{w} V_{wf} + m_{b}V_{bf}

By solving for V_{wf} adn substitute the givens

V_{wf} = \frac{m_{b}V_{bi} - m_{b} V_{bf}    }{m_{W} }

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V_{wf} = 4.67m/s

b) The center of mass speed is defined as

V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}

substituting:

V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)

V = 8.29 m/s

7 0
3 years ago
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