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Lubov Fominskaja [6]
2 years ago
11

A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l

ong line of charge with charge per unit length -6.00 μC/m is parallel to the x-axis at y = 11.0 cm and its midpoint is also at x = 0. At what point on the y -axis is the resultant electric field of the two lines of charge equal to zero?
Physics
1 answer:
artcher [175]2 years ago
9 0

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, \lambda_1=8\times 10^{-6}\ C

linear charge density of the other charge distribution, \lambda_2=-6\times 10^{-6}\ C

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

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

where:

\lambda= linear charge density

r = radial distance from the center of wire

\epsilon_0= permittivity of free space

Therefore,

E_1=E_2

\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}

\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}

\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}

x=0.0629\ m

∴at y=6.29 cm the charge of the two distribution will be equal.

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Answer:

6.67ft/s^2

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Answer:

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

0.010935 J

0.0858675 J

Explanation:

m_1 = Mass of first glider = 0.3 kg

m_2 = Mass of second glider = 0.15 kg

u_1 = Initial Velocity of first glider = 0.8 m/s

u_2 = Initial Velocity of second glider = 0 m/s

v_1 = Final Velocity of first glider

v_2 = Final Velocity of second glider

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s

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The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

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