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An extremely long linear charge distribution, which has a charge density of lambda1 = 4.74 muC/m,...

Question:

An extremely long linear charge distribution, which has a charge density of {eq}\lambda_1 {/eq} = 4.74 {eq}\mu C/m {/eq}, is placed on x-axis. Another very long linear charge distribution lies at y = 0.382 m position as parallel to the x-axis. This second charge distribution has a charge density of {eq}\lambda_2 {/eq} = -5.52 {eq}\mu C/m {/eq}.

Find the magnitude and the direction of the electric field at {eq}Y_P {/eq} = 0.220 m position.

Electric Field Due to an Infinite Line of Charge

Electric field due to an infinite line of charge or a linear charge distribution can be calculated using Gauss law in electrostatics. Gauss law in electrostatics says that the net electric flux {eq}\phi {/eq} through a Gaussian surface enclosing net charge Q is given as {eq}\phi = \dfrac { Q } { \epsilon_0 } {/eq}. Here {eq}\epsilon_0 {/eq} is the permittivity of free space. The electric flux {eq}\phi {/eq} is the dot product of the electric field E through the Gaussian surface and area A of the Gaussian surface.

  • In the case of an infinite line of charge electric field at a radial distance r can be obtained as {eq}E = \dfrac { \lambda } { 2 \pi \epsilon_0 r } {/eq}
  • The linear density of charge {eq}\lambda {/eq} can be defined as the charge available per unit length.
  • Electric field is radially outwards for a positive charge density and radially inwards to the line of charge for a negative charge density.

Answer and Explanation:

Given data

  • Linear charge density of the line of charge lying on the x axis {eq}\lambda_1 = 4.74 \times 10^{-6} \ C/ m {/eq}
  • Linear charge density of the second line of charge passing parallel to the first line {eq}\lambda_2 = - 5.52 \times 10^{-6} \ C/ m {/eq}
  • The distance to the position of the second line of charge from the origin {eq}y_1 = 0.382 \ m {/eq}
  • Distance to the point P from the origin {eq}y = 0.220 \ m {/eq}
  • Permittivity of free space {eq}\epsilon_0 = 8.854 \times 10^{-12} \ F/ m {/eq}

Distance from the second line of charge to the given point {eq}y_2 = y_1 - y \\ y_2 = 0.382 - 0.220 \\ y_2 = 0.162 \ m {/eq}

Magnitude of the electric field at the given point due to the first line of charge {eq}E_1 = \dfrac { \lambda_1 } { 2 \pi \epsilon_0 y } \\ E_1 = \dfrac { 4.74 \times 10^{-6} } { 2 \pi \times 8.854 \times 10^{-12} \times 0.220 } \\ E_1 = 3.87 \times 10^5 \ N/C {/eq}

This electric field is directed along the positive y direction.

Magnitude of the electric field at the given point due to the second line of charge {eq}E_2 = \dfrac { \lambda_2 } { 2 \pi \epsilon_0 y_2 } \\ E_2 = \dfrac { 5.52 \times 10^{-6} } { 2 \pi \times 8.854 \times 10^{-12} \times 0.162 } \\ E_2 = 6.12 \times 10^5 \ N/C {/eq}

This electric field is directed along the positive y axis.

Therefore the net electric field at the given point {eq}E = E_1 + E_2 \\ E = 3.87 \times 10^5 + 6.12 \times 10^5 \\ E = 9.99 \times 10^5 \ N/C {/eq}

The net electric field is directed in the positive y direction.


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