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prohojiy [21]
3 years ago
9

Two very long uniform lines of charge are parallel and are separated by 0.300 m. each line of charge has charge per unit length

+5.20 mc>m. what magnitude of force does one line of charge exert on a 0.0500-m section of the other line of charge?
Physics
1 answer:
Alenkasestr [34]3 years ago
8 0

linear charge density of system of two line charges is given as

\lambda = 5.20 \muC/m

now as we know that electric field due to a line charge at some distance from it is given by

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

so here we will first find the electric field of first line charge at the position of other line charge

E = \frac{5.20 * 10^{-6}}{2 \pi * 8.85 * 10^{-12}* 0.300}

E = 312000 N/C

now as we know that

F = qE

here q = charge on the line charge system at which force is required

E = electric field on that system of charge where force is required

now we can find the charge by

q = \lambda * L

q = 5.20 * 10^{-6}* 0.05 = 0.26 * 10^{-6} C

Now using the above formula

F = qE

F = 0.26 * 10^{-6} * 312000

F = 0.0811 N

so force on the part of wire is F = 0.0811 N

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       v₂ = 0.56 m / s

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b) V_x,i = 44.8 m/s

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d)  V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.  

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

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- V_y,i = 0 (horizontal throw)

                                 55 = 0 + 0 + 0.5*(9.81)*t^2

                                 t = sqrt ( 55 * 2 / 9.81 )

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- Use the second equation of motion in x direction:

                                 x(f) = x(0) + V_x,i*t

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- Use the first equation of motion in y direction:

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                                 V_y,f = 0 + 9.81*3.349

                                 V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

                                 V^2 = V_y,f^2 + V_x,i^2

                                 V = sqrt (32.85^2 + 44.8^2)

                                 V = 55.55 m/s

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